Explain Force Magnitude and Direction - Physics

A $12\,\text{kg}$ crate is pulled across a horizontal floor at constant velocity. The pulling force is $F_{\text{app}}=50\,\text{N}$ to the right (→). The crate experiences kinetic friction $F_f$ to the left (←), weight $F_g$ downward (↓), and normal force $F_N$ upward (↑). Based on this situation, what is the magnitude of the friction force $F_f$ acting on the crate?
Answer: 50 N This question tests understanding of force balance and equilibrium. An object is in equilibrium when the net force is zero, meaning the vector sum of all forces equals zero—for an object at rest or moving at constant velocity, forces must balance in every direction (sum of upward forces = sum of downward forces, sum of rightward forces = sum of leftward forces). In this scenario, the crate is moving at constant velocity, so forces must balance. Vertically: normal force upward ( $F_N$ ) must equal weight downward ( $F_g$ = mg), but the value isn't needed here. Horizontally: applied force (F_app = 50 N) must equal friction force ( $F_f$ ), so $F_f$ = 50 N to the left opposing the rightward applied force. Choice B is correct because it properly applies equilibrium conditions showing forces balance horizontally. Choice D incorrectly assumes the friction force equals the weight ( $F_f$ = mg ≈ 120 N if g=10 m/s²), but friction opposes the applied force and equals it in magnitude for constant velocity, not the weight. When analyzing forces: (1) draw a free body diagram showing all forces on the object, (2) choose a coordinate system and resolve angled forces into components, (3) apply equilibrium conditions (ΣF = 0 in each direction) if object is at rest or constant velocity, or apply F_net = ma if accelerating, and (4) remember that forces in opposite directions subtract while forces in same direction add. Common errors to avoid: (a) assuming normal force always equals weight (it depends on other vertical forces and surface angle), (b) forgetting that friction opposes motion or tendency to move (not just opposes applied force), (c) confusing action-reaction pairs (different objects) with balanced forces (same object), and (d) forgetting to account for all forces when finding net force.
A $6\,\text{kg}$ crate is pulled to the right across a horizontal floor at constant velocity by a horizontal applied force $F_{app}=30\,\text{N}$ . What is the magnitude of the kinetic friction force $F_f$ acting on the crate?
Answer: $30\,\text{N}$ This question tests understanding of force balance and equilibrium for an object moving at constant velocity. An object moving at constant velocity has zero acceleration, which means the net force must be zero according to Newton's First Law—this requires all forces to balance in every direction. In this scenario, the crate moves horizontally at constant velocity, so forces must balance both vertically and horizontally. Vertically: normal force upward ( $F_N$ ) equals weight downward ( $F_g$ = mg = 6 kg × 10 m/s² = 60 N), so forces balance. Horizontally: the applied force F_app = 30 N to the right must be balanced by an equal friction force to the left, so $F_f$ = 30 N opposing the motion. Choice B is correct because it properly applies equilibrium conditions showing that when an object moves at constant velocity, the friction force must exactly balance the applied force. When analyzing forces on objects at constant velocity: (1) recognize that constant velocity means zero acceleration and therefore zero net force, (2) identify all forces and their directions, (3) apply force balance in each direction separately, and (4) remember that kinetic friction opposes the direction of motion with magnitude that depends on the normal force and coefficient of friction, but in equilibrium problems it must equal the applied force.
A $10\,\text{kg}$ box is at rest on a horizontal floor. A student pulls upward on the box with a vertical force $F_{\text{app}}=30\,\text{N}$ (↑), but the box remains in contact with the floor and does not lift. Take $g=10\,\text{m/s}^2$ . Forces on the box: weight $F_g$ (↓), normal force $F_N$ (↑), applied force $F_{\text{app}}$ (↑). What is the magnitude of the normal force $F_N$ ?
Answer: $70\,\text{N}$ This question tests understanding of normal force with additional vertical forces. The normal force is the contact force exerted by a surface perpendicular to the surface, with magnitude that adjusts to prevent the object from passing through the surface—it's not always equal to the object's weight and depends on other forces and surface orientation. In this scenario, the box is at rest and remains in contact, so forces must balance vertically: normal force upward ( $F_N$ ) plus applied force upward (30 N) must equal weight downward ( $F_g$ = 10 kg * 10 m/s² = 100 N), giving $F_N$ = 100 - 30 = 70 N. Choice C is correct because it properly applies equilibrium conditions showing forces balance vertically, with the upward applied force reducing the normal force. Choice A incorrectly adds the applied force to the weight instead of subtracting, but since the applied force is upward, it lightens the effective weight on the surface. When analyzing forces: (1) draw a free body diagram showing all forces on the object, (2) choose a coordinate system and resolve angled forces into components, (3) apply equilibrium conditions (ΣF = 0 in each direction) if object is at rest or constant velocity, or apply F_net = ma if accelerating, and (4) remember that forces in opposite directions subtract while forces in same direction add. Common errors to avoid: (a) assuming normal force always equals weight (it depends on other vertical forces and surface angle), (b) forgetting that friction opposes motion or tendency to move (not just opposes applied force), (c) confusing action-reaction pairs (different objects) with balanced forces (same object), and (d) forgetting to account for all forces when finding net force.
A $6\,\text{kg}$ box is at rest on a horizontal surface. A horizontal pull of $F_{\text{app}}=25\,\text{N}$ acts to the right (→). The coefficient of static friction is $\mu_s=0.50$ . Take $g=10\,\text{m/s}^2$ . What is the magnitude of the static friction force $F_f$ on the box?
Answer: $25\,\text{N}$ This question tests understanding of static friction and equilibrium for an object at rest. An object at rest has zero acceleration, which means the net force must be zero—all forces must balance in every direction, with static friction adjusting to prevent motion. In this scenario, the box remains at rest despite the applied force, so horizontal forces must balance. The maximum possible static friction is $F_f$ ,max = μ_s × $F_N$ = 0.50 × (6 kg × 10 m/s²) = 0.50 × 60 N = 30 N. Since the applied force (25 N) is less than this maximum (30 N), the box doesn't slide, and static friction exactly balances the applied force: $F_f$ = 25 N to the left (←) to oppose the 25 N rightward pull. Choice B (25 N) is correct because static friction adjusts to exactly balance the applied force when the object remains at rest—it's not automatically at its maximum value. Choice C (30 N) incorrectly assumes static friction is always at its maximum value μ_s × $F_N$ , but static friction only reaches this maximum at the threshold of motion—below this threshold, it equals whatever is needed to maintain equilibrium. When analyzing static friction: (1) calculate the maximum possible static friction $F_f$ ,max = μ_s × $F_N$ , (2) compare the applied force to this maximum, (3) if F_app < $F_f$ ,max, the object remains at rest and $F_f$ = F_app, (4) if F_app ≥ $F_f$ ,max, the object begins to slide and kinetic friction takes over. Key insight: static friction is a responsive force that adjusts between 0 and μ_s × $F_N$ to prevent motion.
A $10\,\text{kg}$ box rests on a $20^\circ$ incline. It is held at rest by static friction (no rope). Take $g=10\,\text{m/s}^2$ . The component of weight parallel to the incline points down the slope. In which direction does the static friction force on the box act?
Answer: Up the slope (parallel to the incline) This question tests understanding of force magnitude and direction in multi-force scenarios. When forces act at angles, they must be resolved into perpendicular components using trigonometry: for a force F at angle θ from the horizontal, $F_x$ = F cos(θ) and $F_y$ = F sin(θ), or on an incline the weight component parallel to slope is mg sin(θ) and perpendicular is mg cos(θ). On the incline, we resolve the weight into components: parallel to incline is mg sin(20°) down the slope, and perpendicular is mg cos(20°) into the surface. The normal force equals the perpendicular component, and since the box is at rest, static friction up the slope must equal the parallel component to prevent sliding. Choice B is correct because it identifies the force relationship from force balance or given constraints. Choice A shows friction pointing down the slope, but friction always opposes motion or the tendency to move, so it must point up the slope to oppose the component of weight down the slope. When analyzing forces: (1) draw a free body diagram showing all forces on the object, (2) choose a coordinate system and resolve angled forces into components, (3) apply equilibrium conditions (ΣF = 0 in each direction) if object is at rest or constant velocity, or apply F_net = ma if accelerating, and (4) remember that forces in opposite directions subtract while forces in same direction add. Common errors to avoid: (a) assuming normal force always equals weight (it depends on other vertical forces and surface angle), (b) forgetting that friction opposes motion or tendency to move (not just opposes applied force), (c) confusing action-reaction pairs (different objects) with balanced forces (same object), and (d) forgetting to account for all forces when finding net force.
A $5.0\,\text{kg}$ box rests on a $30^\circ$ incline. The coefficient of kinetic friction is $\mu_k=0.20$ , and the box is sliding down the incline. Take $g=10\,\text{m/s}^2$ . Which direction does the kinetic friction force $F_f$ on the box act?
Answer: Up the incline (parallel to the surface, uphill) This question tests understanding of force magnitude and direction in multi-force scenarios on inclines. When forces act at angles, they must be resolved into perpendicular components using trigonometry: for a force F at angle θ from the horizontal, $F_x$ = F cos(θ) and $F_y$ = F sin(θ), or on an incline the weight component parallel to slope is mg sin(θ) and perpendicular is mg cos(θ). On the incline, we resolve the weight into components: parallel to incline is F_parallel = mg sin(θ) = (5 kg)(10 m/s²) sin(30°) = 25 N down the slope, and perpendicular is F_perp = mg cos(θ) = 43.3 N into the surface. The normal force equals the perpendicular component: $F_N$ ≈ 43.3 N, and since the box is sliding down, kinetic friction acts up the slope to oppose the motion. Choice B is correct because it identifies the force relationship from given constraints, with friction opposing the direction of sliding (down the incline), so it points up the incline. Choice A shows friction pointing in the wrong direction, but friction always opposes motion or the tendency to move, so it must point up the incline to oppose the downhill sliding. When analyzing forces: (1) draw a free body diagram showing all forces on the object, (2) choose a coordinate system and resolve angled forces into components, (3) apply equilibrium conditions (ΣF = 0 in each direction) if object is at rest or constant velocity, or apply F_net = ma if accelerating, and (4) remember that forces in opposite directions subtract while forces in same direction add. Key relationships to remember: on horizontal surface in equilibrium, $F_N$ = $F_g$ and F_app = $F_f$ ; on incline at angle θ, $F_N$ = mg cos(θ) and component down slope = mg sin(θ); for constant velocity motion (net force = 0 in all directions); for accelerating motion (net force ≠ 0 and F_net = ma points in direction of acceleration).
A $10\,\text{kg}$ box rests on a $30^\circ$ incline (angle measured from the horizontal). The coefficient of kinetic friction is $\mu_k=0.20$ , and the box slides down the incline. Using $g=10\,\text{m/s}^2$ , what is the magnitude of the normal force $F_N$ on the box?
Answer: $86.6\,\text{N}$ This question tests understanding of component analysis of angled forces on an inclined plane. When an object rests on an incline, the weight force must be resolved into components: parallel to the incline and perpendicular to the incline—the normal force is the contact force exerted by the surface perpendicular to the surface, equal in magnitude to the perpendicular component of weight when no other forces act perpendicular to the surface. On the incline at 30°, we resolve the weight into components: perpendicular to incline is F_perp = mg cos(θ) = (10 kg)(10 m/s²) cos(30°) = 100 N × 0.866 = 86.6 N into the surface, and parallel to incline is F_parallel = mg sin(θ) = 100 N × 0.5 = 50 N down the slope. The normal force equals the perpendicular component: $F_N$ = 86.6 N, since there are no other forces acting perpendicular to the incline surface. Choice B is correct because it accurately resolves the weight force into components using correct trigonometry—the perpendicular component uses cosine of the incline angle. Choice C incorrectly assumes the normal force equals the full weight ( $F_N$ = mg = 100 N), but this is only true on horizontal surfaces—on an incline, the normal force equals only the perpendicular component of weight, which is mg cos(θ). When analyzing forces on inclines: (1) always resolve weight into components parallel and perpendicular to the surface, (2) use F_perp = mg cos(θ) and F_parallel = mg sin(θ) where θ is measured from horizontal, (3) remember the normal force equals the perpendicular component of weight (not the full weight), and (4) friction acts parallel to the surface opposing motion or tendency to move.
A $6\,\text{kg}$ block is pulled across a horizontal surface by a force $F_{\text{app}}=50\,\text{N}$ at an angle of $37^\circ$ above the horizontal toward the right (↗). Friction acts to the left (←). What is the horizontal component of the applied force, $F_{\text{app},x}$ ? (Use $\cos 37^\circ \approx 0.80$ .)
Answer: 40 N This question tests understanding of component analysis of angled forces. When forces act at angles, they must be resolved into perpendicular components using trigonometry: for a force F at angle θ from the horizontal, $F_x$ = F cos(θ) and $F_y$ = F sin(θ), or on an incline the weight component parallel to slope is mg sin(θ) and perpendicular is mg cos(θ). The applied force of F = 50 N at angle 37° above horizontal has components: horizontal $F_x$ = F cos(37°) ≈ 50 * 0.80 = 40 N rightward, and vertical $F_y$ = F sin(37°) upward, but the question asks only for the horizontal component. Choice B is correct because it accurately resolves force into components using correct trigonometry. Choice D uses sine when it should use cosine (or vice versa) for the component calculation—for a force at angle θ from horizontal, the horizontal component uses cosine: $F_x$ = F cos(θ), not F sin(θ) ≈ 50 * 0.60 = 30 N (choice A) or other miscalculations. When analyzing forces: (1) draw a free body diagram showing all forces on the object, (2) choose a coordinate system and resolve angled forces into components, (3) apply equilibrium conditions (ΣF = 0 in each direction) if object is at rest or constant velocity, or apply F_net = ma if accelerating, and (4) remember that forces in opposite directions subtract while forces in same direction add. Key relationships to remember: on horizontal surface in equilibrium, $F_N$ = $F_g$ and F_app = $F_f$ ; on incline at angle θ, $F_N$ = mg cos(θ) and component down slope = mg sin(θ); for constant velocity motion (net force = 0 in all directions); for accelerating motion (net force ≠ 0 and F_net = ma points in direction of acceleration).
A $20\,\text{kg}$ crate is pushed across a horizontal floor at constant velocity. The coefficient of kinetic friction is $\mu_k=0.30$ , and take $g=10\,\text{m/s}^2$ . The applied force $F_{\text{app}}$ is horizontal. What is the magnitude of the friction force $F_f$ acting on the crate?
Answer: $60\,\text{N}$ This question tests understanding of force balance and equilibrium for an object moving at constant velocity. An object moving at constant velocity has zero acceleration, which means the net force must be zero—all forces must balance in every direction, just as if the object were at rest. In this scenario, the crate moves horizontally at constant velocity, so forces must balance both vertically and horizontally. Vertically: normal force upward ( $F_N$ ) must equal weight downward ( $F_g$ = mg = 20 kg × 10 m/s² = 200 N), giving $F_N$ = 200 N. Horizontally: the applied force F_app to the right must equal the kinetic friction force $F_f$ to the left. Since kinetic friction $F_f$ = μ_k × $F_N$ = 0.30 × 200 N = 60 N, this is the friction force magnitude. Choice B (60 N) is correct because it properly applies the kinetic friction formula $F_f$ = μ_k × $F_N$ with the correct normal force for a horizontal surface. Choice A (20 N) incorrectly uses just the mass value instead of calculating friction, while Choice C (200 N) confuses the normal force with the friction force. When analyzing friction problems: (1) identify whether the object is stationary (static friction) or moving (kinetic friction), (2) find the normal force first (on horizontal surfaces, $F_N$ = mg), (3) calculate friction using $F_f$ = μ × $F_N$ , and (4) remember that for constant velocity motion, friction must equal the applied force to maintain equilibrium.
A $15\,\text{kg}$ box is pushed across a horizontal floor by a force of $F_{\text{app}}=50\,\text{N}$ directed $37^\circ$ below the horizontal (downward and to the right). The box moves at constant velocity. Take $g=10\,\text{m/s}^2$ . What is the magnitude of the normal force $F_N$ on the box?
Answer: $180\,\text{N}$ This question tests understanding of normal force in scenarios with angled forces. The normal force is the contact force exerted by a surface perpendicular to the surface, with magnitude that adjusts to prevent the object from passing through the surface—it's not always equal to the object's weight and depends on other forces and surface orientation. In this scenario, the box is moving at constant velocity, so forces must balance. The applied force of 50 N at 37° below horizontal has components: horizontal $F_x$ = 50 cos(37°) ≈ 40 N rightward, and vertical $F_y$ = 50 sin(37°) ≈ 30 N downward, which adds to the weight. Vertically: normal force upward ( $F_N$ ) must equal weight (150 N) plus downward component (30 N), giving $F_N$ = 180 N. Choice C is correct because it accurately resolves force into components using correct trigonometry and applies equilibrium in the vertical direction. Choice A assumes the normal force always equals the weight ( $F_N$ = mg), but this is only true on horizontal surfaces with no other vertical forces—here the downward angled force increases the effective weight, changing the normal force to $F_N$ = 180 N. When analyzing forces: (1) draw a free body diagram showing all forces on the object, (2) choose a coordinate system and resolve angled forces into components, (3) apply equilibrium conditions (ΣF = 0 in each direction) if object is at rest or constant velocity, or apply F_net = ma if accelerating, and (4) remember that forces in opposite directions subtract while forces in same direction add.
A $10\,\text{kg}$ box rests on a $30^\circ$ incline (angle measured from the horizontal). The surface is frictionless. Take $g=10\,\text{m/s}^2$ . What is the magnitude of the normal force $F_N$ on the box?
Answer: $87\,\text{N}$ This question tests understanding of component analysis of angled forces, specifically the normal force on an inclined plane. When forces act at angles, they must be resolved into perpendicular components using trigonometry: for an object on an incline at angle θ from horizontal, the weight component perpendicular to the surface is mg cos(θ) and parallel to the surface is mg sin(θ). On the incline, we resolve the weight into components: perpendicular to incline is F_perp = mg cos(θ) = (10 kg)(10 m/s²) cos(30°) = 100 N × 0.866 = 86.6 N into the surface, and parallel is F_parallel = mg sin(θ) = 100 N × 0.5 = 50 N down the slope. Since the surface is frictionless and there are no other forces perpendicular to the surface, the normal force equals the perpendicular component: $F_N$ = 86.6 N ≈ 87 N. Choice B (87 N) is correct because it accurately resolves the weight into components using mg cos(30°) for the perpendicular component that determines the normal force. Choice C (100 N) incorrectly assumes the normal force equals the full weight, which is only true on horizontal surfaces—on an incline, the normal force is reduced by the cosine factor. When analyzing forces on inclines: (1) always resolve weight into components parallel and perpendicular to the surface, (2) use mg sin(θ) for the component down the slope and mg cos(θ) for the component into the surface, (3) the normal force equals the perpendicular component when no other perpendicular forces act, and (4) remember that as the incline angle increases, the normal force decreases (becoming zero at 90°).
A $5\,\text{kg}$ box is pulled across a horizontal floor at constant velocity by a rope with tension $F_T=40\,\text{N}$ directed $30^\circ$ above the horizontal. Take $g=10\,\text{m/s}^2$ . What is the magnitude of the normal force $F_N$ on the box?
Answer: $30\,\text{N}$ This question tests understanding of force components and how angled forces affect the normal force. When forces act at angles, they must be resolved into perpendicular components using trigonometry: for a force F at angle θ from the horizontal, $F_x$ = F cos(θ) and $F_y$ = F sin(θ). The rope tension $F_T$ = 40 N at 30° above horizontal has components: horizontal $F_x$ = $F_T$ cos(30°) = 40 N × 0.866 = 34.6 N rightward, and vertical $F_y$ = $F_T$ sin(30°) = 40 N × 0.5 = 20 N upward. For vertical equilibrium (constant velocity means zero acceleration), the upward forces must equal downward forces: $F_N$ + $F_y$ = $F_g$ , so $F_N$ = $F_g$ - $F_y$ = mg - $F_y$ = (5 kg × 10 m/s²) - 20 N = 50 N - 20 N = 30 N. Choice A (30 N) is correct because it properly accounts for the upward component of the tension reducing the normal force from its usual value of mg. Choice B (50 N) incorrectly assumes the normal force equals the weight, ignoring the vertical component of the tension—when an angled force has an upward component, it reduces the normal force below mg. When analyzing angled forces: (1) always resolve into horizontal and vertical components, (2) apply equilibrium conditions separately in each direction, (3) remember that upward force components reduce the normal force while downward components increase it, and (4) the normal force adjusts to maintain vertical equilibrium. Key insight: the normal force is not always equal to weight—it depends on all vertical forces acting on the object.
A $6.0\,\text{kg}$ box rests on a $25^\circ$ incline with no other forces applied. Take $g=10\,\text{m/s}^2$ . Consider the component of the weight parallel to the incline. What is the magnitude of the component of the gravitational force parallel to the incline, $F_{g,\parallel}=mg\sin\theta$ ?
Answer: $60\sin(25^\circ)\,\text{N}\approx 25\,\text{N}$ This question tests understanding of component analysis of angled forces on inclines. When forces act at angles, they must be resolved into perpendicular components using trigonometry: for a force F at angle θ from the horizontal, $F_x$ = F cos(θ) and $F_y$ = F sin(θ), or on an incline the weight component parallel to slope is mg sin(θ) and perpendicular is mg cos(θ). On the incline, we resolve the weight into components: parallel to incline is F_parallel = mg sin(θ) = (6 kg)(10 m/s²) sin(25°) ≈ 60 * 0.4226 ≈ 25 N down the slope, and perpendicular is F_perp = mg cos(θ) ≈ 54 N into the surface. Choice A is correct because it accurately resolves the gravitational force into components using correct trigonometry, with $F_{g,parallel}$ = mg sinθ ≈ 25 N. Choice B uses sine when it should use cosine (or vice versa) for the component calculation—for the parallel component on an incline, it's mg sin(θ), not mg cos(θ). When analyzing forces: (1) draw a free body diagram showing all forces on the object, (2) choose a coordinate system and resolve angled forces into components, (3) apply equilibrium conditions (ΣF = 0 in each direction) if object is at rest or constant velocity, or apply F_net = ma if accelerating, and (4) remember that forces in opposite directions subtract while forces in same direction add. Key relationships to remember: on horizontal surface in equilibrium, $F_N$ = $F_g$ and F_app = $F_f$ ; on incline at angle θ, $F_N$ = mg cos(θ) and component down slope = mg sin(θ); for constant velocity motion (net force = 0 in all directions); for accelerating motion (net force ≠ 0 and F_net = ma points in direction of acceleration).
A $10\,\text{kg}$ sled is pulled on level snow by a horizontal rope force $F_{\text{app}}=80\,\text{N}$ to the right (→). The coefficient of kinetic friction is $\mu_k=0.30$ . Take $g=10\,\text{m/s}^2$ . Forces: $F_g$ (↓), $F_N$ (↑), $F_{\text{app}}$ (→), $F_f$ (←). What is the net horizontal force on the sled (magnitude and direction)?
Answer: $50\,\text{N}$ to the right (→) This question tests understanding of net force determination in multi-force scenarios. Newton's Second Law (F_net = ma) states that the net force on an object equals its mass times acceleration—when forces don't balance, the object accelerates in the direction of the net force with magnitude a = F_net / m. The forces acting on the sled are: weight 100 N downward, normal force 100 N upward, applied force 80 N to the right, and friction $F_f$ = μ $F_N$ = 0.30 * 100 = 30 N to the left. The net vertical force is 0 N (balanced). The net horizontal force is 80 - 30 = 50 N to the right. Choice C is correct because it correctly calculates net force as vector sum with directions, considering opposing forces subtract. Choice D calculates an individual force magnitude when the question asks for net force—net force requires vector addition considering directions: F_net = F_right - F_left, not just the magnitude of one force. When analyzing forces: (1) draw a free body diagram showing all forces on the object, (2) choose a coordinate system and resolve angled forces into components, (3) apply equilibrium conditions (ΣF = 0 in each direction) if object is at rest or constant velocity, or apply F_net = ma if accelerating, and (4) remember that forces in opposite directions subtract while forces in same direction add. Common errors to avoid: (a) assuming normal force always equals weight (it depends on other vertical forces and surface angle), (b) forgetting that friction opposes motion or tendency to move (not just opposes applied force), (c) confusing action-reaction pairs (different objects) with balanced forces (same object), and (d) forgetting to account for all forces when finding net force.
A $10\,\text{kg}$ box is pulled across a horizontal floor by an applied force $F_{\text{app}}=60\,\text{N}$ at $30^\circ$ above the horizontal (to the right). Assume $g=10\,\text{m/s}^2$ . Which expression best gives the normal force magnitude $F_N$ on the box (assuming the box does not accelerate vertically)?
Answer: $F_N = mg - F_{\text{app}}\sin 30^\circ$ This question tests understanding of relationships between force types. The normal force is the contact force exerted by a surface perpendicular to the surface, with magnitude that adjusts to prevent the object from passing through the surface—it's not always equal to the object's weight and depends on other forces and surface orientation. The applied force of F_app = 60 N at angle 30° above horizontal has components: horizontal $F_x$ = F_app cos(30°) rightward, and vertical $F_y$ = F_app sin(30°) upward, which affects the normal force making $F_N$ = mg - $F_y$ = mg - F_app sin(30°), since no vertical acceleration. Vertically, forces balance: $F_N$ + F_app sin(30°) = mg. Choice B is correct because it correctly determines force magnitude from force balance or given constraints. Choice D assumes the normal force always equals the weight ( $F_N$ = mg), but this is only true on horizontal surfaces with no other vertical forces—here the upward component of the applied force reduces the normal force to $F_N$ = mg - F_app sin(30°). When analyzing forces: (1) draw a free body diagram showing all forces on the object, (2) choose a coordinate system and resolve angled forces into components, (3) apply equilibrium conditions (ΣF = 0 in each direction) if object is at rest or constant velocity, or apply F_net = ma if accelerating, and (4) remember that forces in opposite directions subtract while forces in same direction add. Common errors to avoid: (a) assuming normal force always equals weight (it depends on other vertical forces and surface angle), (b) forgetting that friction opposes motion or tendency to move (not just opposes applied force), (c) confusing action-reaction pairs (different objects) with balanced forces (same object), and (d) forgetting to account for all forces when finding net force.
A $15\,\text{kg}$ box rests on a $20^\circ$ incline (angle from the horizontal) and is held at rest by static friction only (no rope). Using $g=10\,\text{m/s}^2$ , what is the required magnitude of the static friction force $F_f$ (along the incline) to keep the box in equilibrium?
Answer: $51\,\text{N}$ This question tests understanding of force balance on an inclined plane in static equilibrium. An object is in equilibrium when the net force is zero—for an object at rest on an incline held by friction alone, the static friction force must exactly balance the component of weight parallel to the incline. On the incline at 20°, we resolve the weight into components: parallel to incline is F_parallel = mg sin(θ) = (15 kg)(10 m/s²) sin(20°) = 150 N × 0.342 = 51.3 N down the slope, and perpendicular is F_perp = mg cos(θ) = 150 N × 0.940 = 141 N into the surface. Since the box is at rest (equilibrium), the static friction force up the incline must equal the parallel component of weight: $F_f$ = 51.3 N ≈ 51 N up the slope. Choice C is correct because it accurately calculates the parallel component of weight that friction must balance for equilibrium on the incline. Choice B incorrectly identifies the perpendicular component (141 N) as the friction force, but friction acts parallel to the surface, not perpendicular—the perpendicular component determines the normal force, not friction. When analyzing static equilibrium on inclines: (1) resolve weight into components parallel and perpendicular to the surface, (2) recognize that static friction adjusts to balance the parallel component of weight (up to its maximum value), (3) use F_parallel = mg sin(θ) for the component friction must balance, and (4) remember friction acts parallel to the surface, opposing the tendency to slide.
A $12\,\text{kg}$ crate is pulled across a horizontal floor at constant velocity by a rope that exerts an applied force $F_{\text{app}}=60\,\text{N}$ to the right (→). The forces on the crate are: weight $F_g$ downward (↓), normal force $F_N$ upward (↑), applied force $F_{\text{app}}$ to the right (→), and kinetic friction $F_f$ to the left (←). What is the magnitude of the friction force $F_f$ on the crate?
Answer: $60\,\text{N}$ This question tests understanding of force balance and equilibrium in multi-force scenarios. An object is in equilibrium when the net force is zero, meaning the vector sum of all forces equals zero—for an object at rest or moving at constant velocity, forces must balance in every direction (sum of upward forces = sum of downward forces, sum of rightward forces = sum of leftward forces). In this scenario, the crate is moving at constant velocity, so forces must balance. Vertically: normal force upward ( $F_N$ ) must equal weight downward ( $F_g$ = 12 kg * 10 m/s² = 120 N), giving $F_N$ = 120 N. Horizontally: applied force (F_app = 60 N) must equal friction force ( $F_f$ ), so $F_f$ = 60 N to the left opposing the rightward applied force. Choice C is correct because it properly applies equilibrium conditions showing forces balance horizontally, with $F_f$ = F_app = 60 N. Choice A incorrectly assumes the friction is equal to the mass or some unrelated value, but friction adjusts to match the applied force in horizontal equilibrium. When analyzing forces: (1) draw a free body diagram showing all forces on the object, (2) choose a coordinate system and resolve angled forces into components, (3) apply equilibrium conditions (ΣF = 0 in each direction) if object is at rest or constant velocity, or apply F_net = ma if accelerating, and (4) remember that forces in opposite directions subtract while forces in same direction add. Common errors to avoid: (a) assuming normal force always equals weight (it depends on other vertical forces and surface angle), (b) forgetting that friction opposes motion or tendency to move (not just opposes applied force), (c) confusing action-reaction pairs (different objects) with balanced forces (same object), and (d) forgetting to account for all forces when finding net force.
A $10\,\text{kg}$ box is on a horizontal surface. A student pushes it with a horizontal force $F_{\text{app}}=30\,\text{N}$ to the right (→), but the box does not move. The coefficient of static friction is $\mu_s=0.50$ and $g=10\,\text{m/s}^2$ . What is the magnitude of the static friction force $F_f$ on the box?
Answer: 30 N This question tests understanding of force balance and equilibrium. An object is in equilibrium when the net force is zero, meaning the vector sum of all forces equals zero—for an object at rest or moving at constant velocity, forces must balance in every direction. In this scenario, the box is at rest (does not move), so forces must balance. Vertically: normal force upward ( $F_N$ ) must equal weight downward ( $F_g$ = 100 N), giving $F_N$ = 100 N. Horizontally: applied force (F_app = 30 N) must equal static friction force ( $F_f$ ), so $F_f$ = 30 N to the left opposing the rightward applied force, since 30 N < maximum static friction (μ_s $F_N$ = 0.50 * 100 = 50 N). Choice C is correct because it properly applies equilibrium conditions showing forces balance. Choice D calculates the maximum static friction (μ_s $F_N$ = 50 N) when the question asks for the actual friction force—since the box doesn't move and applied force is less than max, static friction equals the applied force, not the maximum. When analyzing forces: (1) draw a free body diagram showing all forces on the object, (2) choose a coordinate system and resolve angled forces into components, (3) apply equilibrium conditions (ΣF = 0 in each direction) if object is at rest or constant velocity, or apply F_net = ma if accelerating, and (4) remember that forces in opposite directions subtract while forces in same direction add. Common errors to avoid: (a) assuming normal force always equals weight (it depends on other vertical forces and surface angle), (b) forgetting that friction opposes motion or tendency to move (not just opposes applied force), (c) confusing action-reaction pairs (different objects) with balanced forces (same object), and (d) forgetting to account for all forces when finding net force.
A sled of mass $25\,\text{kg}$ is pulled on level snow by a horizontal rope force $F_T=120\,\text{N}$ to the right (→). The kinetic friction force has magnitude $F_f=45\,\text{N}$ to the left (←). Take $g=10\,\text{m/s}^2$ . What is the magnitude of the net force on the sled?
Answer: $75\,\text{N}$ This question tests understanding of net force determination when multiple horizontal forces act on an object. Newton's Second Law (F_net = ma) states that the net force on an object equals its mass times acceleration—when forces don't balance, the object accelerates in the direction of the net force with magnitude a = F_net / m. The forces acting on the sled are: tension force $F_T$ = 120 N to the right (→) and friction force $F_f$ = 45 N to the left (←). Since these forces act in opposite directions, we must subtract them to find the net force. The net horizontal force is F_net = $F_T$ - $F_f$ = 120 N - 45 N = 75 N to the right, so the sled accelerates rightward with net force magnitude 75 N. Choice C (75 N) is correct because it properly calculates net force as the vector sum, subtracting the opposing friction force from the applied tension force. Choice B (45 N) shows only the friction force magnitude, ignoring the tension force, while Choice D (165 N) incorrectly adds the forces instead of subtracting them—forces in opposite directions must subtract, not add. When calculating net force: (1) identify all forces acting on the object and their directions, (2) forces in the same direction add while forces in opposite directions subtract, (3) the net force direction is toward whichever side has the larger total force, and (4) use F_net = ma to find acceleration if needed. Common errors include forgetting to account for direction (treating all forces as positive) or calculating individual force magnitudes when the question asks for net force.
A $4\,\text{kg}$ cart moves to the right (→) on a horizontal track. A constant friction force $F_f=6\,\text{N}$ acts to the left (←). A motor provides an applied force $F_{\text{app}}=14\,\text{N}$ to the right (→). Weight and normal force cancel vertically. What is the cart's acceleration (magnitude and direction)?
Answer: $2.0\,\text{m/s}^2$ to the right (→) This question tests understanding of net force determination. Newton's Second Law (F_net = ma) states that the net force on an object equals its mass times acceleration—when forces don't balance, the object accelerates in the direction of the net force with magnitude a = F_net / m. The forces acting on the cart are: applied force 14 N to the right, friction 6 N to the left; weight and normal force cancel vertically. The net horizontal force is F_app - $F_f$ = 14 - 6 = 8 N to the right, so acceleration a = 8 N / 4 kg = 2.0 m/s² to the right. Choice B is correct because it correctly determines force magnitude from Newton's Second Law. Choice D calculates an individual force magnitude when the question asks for net force—net force requires vector addition considering directions: F_net = F_right - F_left, not just the magnitude of one force. When analyzing forces: (1) draw a free body diagram showing all forces on the object, (2) choose a coordinate system and resolve angled forces into components, (3) apply equilibrium conditions (ΣF = 0 in each direction) if object is at rest or constant velocity, or apply F_net = ma if accelerating, and (4) remember that forces in opposite directions subtract while forces in same direction add. Key relationships to remember: on horizontal surface in equilibrium, $F_N$ = $F_g$ and F_app = $F_f$ ; on incline at angle θ, $F_N$ = mg cos(θ) and component down slope = mg sin(θ); for constant velocity motion (net force = 0 in all directions); for accelerating motion (net force ≠ 0 and F_net = ma points in direction of acceleration).

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