Forces and Free-Body Diagrams - AP Physics C Mechanics

A box sits on the bed of a pickup truck that is accelerating forward. The box does not slide relative to the truck bed. In the free-body diagram for the box, which force provides the acceleration, and what can be concluded about its magnitude?
Answer: Static friction from the truck bed provides acceleration, and its magnitude equals $ma$ in the forward direction When analyzing forces on objects that move together without slipping, you need to identify which force actually causes the acceleration by applying Newton's second law to each object separately. Since the box doesn't slide relative to the truck bed, both have the same acceleration $a$ forward. For the box to accelerate forward, there must be a net horizontal force acting on it. Looking at the box's free-body diagram, the only horizontal force available is static friction from the truck bed surface. This friction force points forward (in the direction of acceleration) and must equal $ma$ to satisfy Newton's second law: $F_{net} = ma$ . Choice A incorrectly identifies the normal force as providing acceleration. The normal force acts vertically upward, balancing the box's weight. Even if the truck bed were tilted, any horizontal component of the normal force alone couldn't account for the full acceleration. Choice B misunderstands the physics entirely. Air resistance opposes motion and would act backward on the box, not forward. Air doesn't "push" objects forward as vehicles move through it. Choice C confuses the system analysis. The engine accelerates the truck, and the truck then accelerates the box through friction. There's no direct mechanical connection transmitting engine force to the box. Remember this key principle: when objects accelerate together without slipping, look for the contact force between them that prevents relative motion. That contact force (usually friction) is what accelerates the "passenger" object. This applies to boxes on trucks, people in elevators, or any situation where objects move together.
A car is traveling at a constant speed as it rounds a flat, unbanked circular curve. What is the primary purpose of drawing a free-body diagram for the car in this situation?
Answer: To identify the real forces whose vector sum provides the net centripetal force. A free-body diagram is used to identify all the real, physical forces acting on an object. For an object in uniform circular motion, there must be a net force directed towards the center of the circle (the centripetal force). The FBD helps identify which force or combination of forces (in this case, static friction) provides this required net force. Centrifugal force is a fictitious force, and the net force is not zero because the direction of velocity is changing.
A book rests motionless on a horizontal tabletop. Which statement correctly identifies the forces that must be included on a complete free-body diagram for the book?
Answer: A downward gravitational force exerted by the Earth and an upward normal force exerted by the tabletop. A free-body diagram shows all external forces acting on the object. The Earth exerts a downward gravitational force. Since the book is in contact with the tabletop and is not accelerating vertically, the tabletop must exert an equal and opposite upward force, which is the normal force. These are the only two forces acting on the book in this situation.
A passenger is in an elevator that is accelerating downwards, but at a rate less than $g$ . A free-body diagram of the passenger should show which of the following?
Answer: A downward gravitational force and an upward normal force, with the gravitational force having a greater magnitude than the normal force. The two forces on the passenger are gravity (down) and the normal force from the elevator floor (up). According to Newton's second law, the net force must be in the direction of acceleration. Since the passenger is accelerating downwards, the net force must be downwards. This means the downward gravitational force must be greater in magnitude than the upward normal force.
A block hangs at rest from two strings of unequal length, which are attached to the ceiling at different points. The block is not centered between the attachment points. What must be true about the forces on the free-body diagram for the block?
Answer: The horizontal components of the two tension forces must be equal in magnitude and opposite in direction. Since the block is in static equilibrium, the net force in any direction must be zero. Analyzing the horizontal forces, the horizontal component of tension from one string must exactly balance the horizontal component of tension from the other string. Because the angles are likely different, the magnitudes of the tensions will not be equal, and their vertical components will sum to balance the gravitational force but are not necessarily equal to each other.
An object is being pulled up a rough inclined plane at a constant velocity by a rope parallel to the plane. Which of the following sets of forces should be shown on a correct free-body diagram for the object?
Answer: Tension, gravity, normal force, and kinetic friction, with the net force equal to zero. The forces acting on the object are: tension from the rope (up the incline), gravity (straight down), the normal force from the plane (perpendicular to the surface), and kinetic friction (down the incline, opposing motion). Since the velocity is constant, the acceleration is zero, and thus the net force must be zero. The friction is kinetic because the object is moving.
A 2.0 kg block rests on top of a 3.0 kg block, which sits on a horizontal frictionless surface. The coefficient of static friction between the blocks is $\mu_s = 0.25$ . A horizontal force $F$ is applied to the bottom block. In the free-body diagram analysis, what is the maximum value of $F$ that can be applied before the blocks begin to slide relative to each other? (Use $g = 10 \text{ m/s}^2$ )
Answer: $F = 12.5$ N, determined by the maximum static friction between the blocks The maximum static friction between blocks is $f_{max} = \mu_s m_{top} g = 0.25 × 2.0 × 10 = 5.0$ N. This friction must accelerate the top block: $f = m_{top} a$ , so $a_{max} = \frac{5.0}{2.0} = 2.5$ m/s². For both blocks to accelerate together at this rate: $F = (m_{top} + m_{bottom}) a_{max} = (2.0 + 3.0) × 2.5 = 12.5$ N. Choice B only considers the friction force magnitude, not the total system. Choice C neglects the top block's mass. Choice D incorrectly relates force directly to weight and friction coefficient.
A uniform ladder leans against a smooth, vertical wall while resting on a rough, horizontal floor. The ladder is in static equilibrium. The free-body diagram for the ladder must include which combination of forces?
Answer: Gravity, a normal force from the wall, a normal force from the floor, and a static friction force from the floor. The forces acting on the ladder are: its weight (gravity) acting at its center of mass, a normal force from the floor directed upward, a normal force from the wall directed horizontally away from the wall, and a static friction force from the floor directed horizontally toward the wall. The wall is smooth, so it exerts no friction. The friction from the floor is necessary to prevent the base of the ladder from slipping away from the wall.
A 2.0 kg block rests on a horizontal surface with coefficient of static friction $\mu_s = 0.40$ and coefficient of kinetic friction $\mu_k = 0.30$ . A horizontal force $F$ is applied to the block, starting from zero and gradually increasing. When analyzing the free-body diagram at the moment just before the block begins to slide, which statement about the forces is correct?
Answer: The friction force equals the applied force $F$ in magnitude and acts opposite to its direction Just before sliding begins, the block is still in static equilibrium, so the net force must be zero. The static friction force adjusts to exactly balance the applied force, up to its maximum value of $\mu_s mg$ . At the moment just before sliding, $F < \mu_s mg$ , so the friction force equals $F$ in magnitude. Choice A is incorrect because the friction force only equals $\mu_s mg$ at the threshold of sliding. Choice C is wrong because kinetic friction hasn't begun yet. Choice D is incorrect because a horizontal force doesn't affect the normal force on a horizontal surface.
A car travels around a horizontal circular curve of radius $R$ at constant speed $v$ . The road surface provides the centripetal force through friction. In the free-body diagram for the car, which statement correctly describes the direction and magnitude of the friction force?
Answer: The friction force points radially inward toward the center with magnitude $\frac{mv^2}{R}$ to provide centripetal acceleration For circular motion at constant speed, the car needs centripetal force $F_c = \frac{mv^2}{R}$ directed toward the center of the circle. Friction provides this force, so it points radially inward with magnitude $\frac{mv^2}{R}$ . Choice A incorrectly describes kinetic friction opposing tangential motion. Choice C wrongly suggests friction balances a centrifugal force (which doesn't exist in an inertial frame). Choice D misunderstands that constant speed still requires centripetal acceleration and force.
The purpose of a free-body diagram is to isolate an object and consider only the external forces acting upon it. Which of the following is an example of an internal force that would NOT be shown on the free-body diagram of a car?
Answer: The force exerted by the car's engine on its transmission. Internal forces are forces that parts of a system exert on each other. The force of the engine on the transmission is internal to the car as a system. External forces are exerted on the system by objects in its environment. Gravity (from Earth), friction (from the road), and air resistance (from the air) are all external forces and would be included on the free-body diagram of the car.
A box is sliding down a frictionless ramp. Which of the following correctly describes the forces that should be on the box's free-body diagram?
Answer: A gravitational force and a normal force, where the normal force is smaller in magnitude than the gravitational force. The forces on the box are gravity (acting vertically downward) and the normal force (acting perpendicular to the ramp's surface). Since the gravitational force has a component perpendicular to the ramp that is balanced by the normal force, the normal force must be smaller in magnitude than the total gravitational force ( $F_N = mg \cos{\theta}$ ). There is no 'force of motion'; the motion is a result of the net force.
A block is pulled up a rough inclined plane at constant velocity by a force $\vec{F}$ applied parallel to the incline. The incline makes angle $\theta$ with the horizontal, and the coefficient of kinetic friction is $\mu_k$ . In the free-body diagram analysis, which expression correctly represents the relationship between the applied force and other forces?
Answer: $F = mg\sin\theta + \mu_k mg\cos\theta$ because friction opposes motion up the incline At constant velocity, the net force is zero, so the applied force must balance both the component of weight down the incline ( $mg\sin\theta$ ) and the friction force opposing motion ( $\mu_k N = \mu_k mg\cos\theta$ ). Both oppose the applied force, so $F = mg\sin\theta + \mu_k mg\cos\theta$ . Choice B incorrectly subtracts friction, suggesting it aids upward motion. Choice C confuses sine and cosine components. Choice D has the wrong sign relationship and would only apply if friction somehow overcame gravity.
A block is pulled by a rope at an angle $\theta$ above the horizontal across a rough surface, causing the block to accelerate. How does the magnitude of the normal force $F_N$ compare to the magnitude of the gravitational force $mg$ ?
Answer: $F_N < mg$ The free-body diagram includes gravity (down), tension (at angle $\theta$ ), normal force (up), and friction (horizontal). Analyzing the vertical forces, the upward normal force plus the upward component of tension must balance the downward gravitational force. So, $F_N + T \sin{\theta} = mg$ . Since $T \sin{\theta}$ is a positive value, $F_N$ must be less than $mg$ .
A $2.0\,\text{kg}$ block rests on a $40^\circ$ incline with $\mu_s=0.50$ and no applied force. Consider the scenario described above, with the block initially at rest and $g=9.8\,\text{m/s}^2$ . Static friction acts along the plane as needed, up to $\mu_s N$ . The normal force is perpendicular to the plane and weight is vertical. Is the object in equilibrium? Justify your answer.
Answer: No; $mg\sin\theta > \mu_s mg\cos\theta$ This question tests AP Physics C: Mechanics skills, specifically understanding forces and free-body diagrams. Free-body diagrams on inclined planes require careful analysis of static friction, which can vary from zero up to μₛN depending on what's needed to prevent motion. In this scenario, a 2.0 kg block rests on a 40° incline with static friction coefficient 0.50, and we must determine if it remains in equilibrium. Choice B is correct because the component of weight down the plane is mg sin(40°) = 12.6 N, while the maximum static friction is μₛmg cos(40°) = 0.50(2.0)(9.8)cos(40°) = 7.5 N; since 12.6 N > 7.5 N, static friction cannot balance the weight component, so the block will slide. Choice A is incorrect because it reverses the inequality - students often confuse when to use sine versus cosine or compare the forces incorrectly. To help students: Teach the critical angle concept where tan(θ) = μₛ determines the steepest angle for equilibrium, and practice comparing the weight component down the plane with maximum static friction. Emphasize that static friction adjusts to match the applied force up to its maximum value. Watch for: students always using μₛN for static friction instead of recognizing it as a maximum, confusing the conditions for static equilibrium, or mixing up force components.
Two masses $m_1=2.0\,\text{kg}$ and $m_2=2.5\,\text{kg}$ are connected by a light rope over a frictionless pulley. Consider the scenario described above, with $m_2$ moving downward and the system accelerating. Take $g=9.8\,\text{m/s}^2$ and assume the rope is massless. Tension is the same on both sides. Determine the tension in the rope.
Answer: $21.8\,\text{N}$ This question tests AP Physics C: Mechanics skills, specifically understanding forces and free-body diagrams. Free-body diagrams for Atwood machines show that tension is the same throughout a massless rope, and both masses experience the same magnitude of acceleration in opposite directions. In this scenario, masses of 2.0 kg and 2.5 kg are connected over a frictionless pulley, with the heavier mass accelerating downward. Choice D is correct because the system acceleration is a = (m₂ - m₁)g/(m₁ + m₂) = (0.5)(9.8)/4.5 = 1.09 m/s², and the tension T = m₁(g + a) = 2.0(9.8 + 1.09) = 21.8 N, which also equals m₂(g - a) as a check. Choice E is incorrect because it represents the weight of the lighter mass (19.6 N), showing the common error of assuming tension equals weight when the system accelerates. To help students: Set up separate free-body diagrams and equations for each mass, use the constraint that both accelerations have the same magnitude, and solve the system of equations. Verify the answer by checking that the same tension satisfies both force equations. Watch for: students assuming tension equals one of the weights, using different accelerations for each mass, or making sign errors in the force equations.
A student is analyzing the forces on a block on an inclined plane. According to the course framework, which convention should be followed when drawing the forces on the free-body diagram?
Answer: Draw all force vectors originating from the dot representing the object, without showing components. The standard convention for AP Physics free-body diagrams is to represent the object as a dot and draw all individual external forces as vectors originating from that dot. Force components are used for calculations but should not be drawn on the primary free-body diagram itself. The net force is the vector sum of the individual forces and is also not drawn on the diagram.
A projectile is launched at an angle and is at the highest point of its trajectory. Neglecting air resistance, which forces should be included on the free-body diagram of the projectile at this instant?
Answer: Only the downward force of gravity. Once a projectile is launched, and neglecting air resistance, the only force acting on it throughout its flight is the gravitational force exerted by the Earth, which always acts straight down. There is no force propelling it forward after it has left the launcher. Its vertical velocity is momentarily zero at the peak, but this does not mean the net force is zero; it is still accelerating downwards due to gravity.
Two blocks, $M_1$ and $M_2$ , are in contact on a frictionless horizontal surface. A horizontal force $F$ is applied to $M_1$ , causing both blocks to accelerate. What forces should be shown on a free-body diagram for block $M_2$ ?
Answer: Gravity, the normal force from the surface, and a contact force exerted by $M_1$ on $M_2$ . A free-body diagram shows forces exerted on a specific object. For block $M_2$ , the forces are: gravity (down), the normal force from the horizontal surface (up), and the contact force from block $M_1$ pushing it (horizontally). The applied force $F$ acts directly on $M_1$ , not $M_2$ .
A block is at rest on a rough inclined plane. A student correctly draws a free-body diagram for the block. Which statement is a defining characteristic of this diagram?
Answer: The vector sum of all forces shown on the diagram is zero. The block is at rest, which means it is in static equilibrium. A key principle of static equilibrium is that the net force on the object is zero. Therefore, the vector sum of all forces (gravity, normal force, and static friction) shown on the free-body diagram must be zero.

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