Explain Force Magnitude and Direction - Physics
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What is the SI unit and common symbol for force magnitude?
What is the SI unit and common symbol for force magnitude?
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Newton, $\text{N}$. Force is measured in newtons, the SI unit for force.
Newton, $\text{N}$. Force is measured in newtons, the SI unit for force.
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What does the direction of the net force indicate about an object’s acceleration?
What does the direction of the net force indicate about an object’s acceleration?
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Acceleration is in the same direction as $\vec{F}_{\text{net}}$. Newton's second law: force causes acceleration in its direction.
Acceleration is in the same direction as $\vec{F}_{\text{net}}$. Newton's second law: force causes acceleration in its direction.
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State Newton’s second law in vector form relating net force and acceleration.
State Newton’s second law in vector form relating net force and acceleration.
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$\vec{F}_{\text{net}} = m\vec{a}$. Vector form shows force and acceleration point the same way.
$\vec{F}_{\text{net}} = m\vec{a}$. Vector form shows force and acceleration point the same way.
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What is the condition for translational equilibrium in terms of net force?
What is the condition for translational equilibrium in terms of net force?
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$\vec{F}_{\text{net}} = \vec{0}$. No net force means no acceleration (equilibrium).
$\vec{F}_{\text{net}} = \vec{0}$. No net force means no acceleration (equilibrium).
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What is the relationship between force magnitude and acceleration when mass is constant?
What is the relationship between force magnitude and acceleration when mass is constant?
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$a \propto F_{\text{net}}$. Direct proportionality: doubling force doubles acceleration.
$a \propto F_{\text{net}}$. Direct proportionality: doubling force doubles acceleration.
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What is the relationship between mass and acceleration when net force is constant?
What is the relationship between mass and acceleration when net force is constant?
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$a \propto \frac{1}{m}$. Inverse relationship: doubling mass halves acceleration.
$a \propto \frac{1}{m}$. Inverse relationship: doubling mass halves acceleration.
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What is the weight force on a mass $m$ near Earth in magnitude and direction?
What is the weight force on a mass $m$ near Earth in magnitude and direction?
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$W = mg$, downward. Weight equals mass times gravitational acceleration.
$W = mg$, downward. Weight equals mass times gravitational acceleration.
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For an object on a horizontal surface with no vertical acceleration, how do $N$ and $mg$ compare?
For an object on a horizontal surface with no vertical acceleration, how do $N$ and $mg$ compare?
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$N = mg$. Normal force balances weight when there's no vertical motion.
$N = mg$. Normal force balances weight when there's no vertical motion.
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What is the direction of the normal force exerted by a surface on an object?
What is the direction of the normal force exerted by a surface on an object?
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Perpendicular to the surface, away from the surface. Normal force pushes outward from the contact surface.
Perpendicular to the surface, away from the surface. Normal force pushes outward from the contact surface.
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What is the direction of kinetic friction relative to the direction of motion?
What is the direction of kinetic friction relative to the direction of motion?
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Opposite the direction of relative motion. Friction opposes sliding between surfaces.
Opposite the direction of relative motion. Friction opposes sliding between surfaces.
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State the magnitude formula for kinetic friction on a surface with coefficient $\mu_k$.
State the magnitude formula for kinetic friction on a surface with coefficient $\mu_k$.
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$f_k = \mu_k N$. Kinetic friction equals coefficient times normal force.
$f_k = \mu_k N$. Kinetic friction equals coefficient times normal force.
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State the inequality for the magnitude of static friction with coefficient $\mu_s$.
State the inequality for the magnitude of static friction with coefficient $\mu_s$.
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$f_s \le \mu_s N$. Static friction adjusts up to maximum $\mu_s N$.
$f_s \le \mu_s N$. Static friction adjusts up to maximum $\mu_s N$.
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What is the direction of static friction when an applied force tends to move an object rightward?
What is the direction of static friction when an applied force tends to move an object rightward?
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Leftward (opposes the impending relative motion). Static friction prevents motion in the applied direction.
Leftward (opposes the impending relative motion). Static friction prevents motion in the applied direction.
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What is the direction of the tension force exerted by a taut rope on an attached object?
What is the direction of the tension force exerted by a taut rope on an attached object?
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Along the rope, away from the object. Tension pulls along the rope's direction.
Along the rope, away from the object. Tension pulls along the rope's direction.
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What does Newton’s third law state about the magnitudes and directions of an action–reaction pair?
What does Newton’s third law state about the magnitudes and directions of an action–reaction pair?
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Equal magnitudes, opposite directions, on different objects. Forces come in equal-opposite pairs on different objects.
Equal magnitudes, opposite directions, on different objects. Forces come in equal-opposite pairs on different objects.
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Identify the net force direction if two forces act: $10,\text{N}$ right and $6,\text{N}$ left.
Identify the net force direction if two forces act: $10,\text{N}$ right and $6,\text{N}$ left.
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$4,\text{N}$ right. Net force is the vector sum: $10-6=4$ rightward.
$4,\text{N}$ right. Net force is the vector sum: $10-6=4$ rightward.
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Identify $\vec{F}_{\text{net}}$ if forces are $8,\text{N}$ up and $8,\text{N}$ down.
Identify $\vec{F}_{\text{net}}$ if forces are $8,\text{N}$ up and $8,\text{N}$ down.
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$0,\text{N}$ (no net force). Equal opposite forces cancel to zero net force.
$0,\text{N}$ (no net force). Equal opposite forces cancel to zero net force.
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Find the acceleration magnitude and direction if $m=2,\text{kg}$ and $\vec{F}_{\text{net}}=6,\text{N}$ left.
Find the acceleration magnitude and direction if $m=2,\text{kg}$ and $\vec{F}_{\text{net}}=6,\text{N}$ left.
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$3,\text{m/s}^2$ left. Using $a=F/m$: $6,\text{N}/2,\text{kg}=3,\text{m/s}^2$.
$3,\text{m/s}^2$ left. Using $a=F/m$: $6,\text{N}/2,\text{kg}=3,\text{m/s}^2$.
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Find $f_k$ if $\mu_k=0.20$ and $N=50,\text{N}$.
Find $f_k$ if $\mu_k=0.20$ and $N=50,\text{N}$.
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$10,\text{N}$. Kinetic friction: $f_k=\mu_k N=0.20×50=10,\text{N}$.
$10,\text{N}$. Kinetic friction: $f_k=\mu_k N=0.20×50=10,\text{N}$.
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Find the weight magnitude for $m=5,\text{kg}$ using $g=9.8,\text{m/s}^2$.
Find the weight magnitude for $m=5,\text{kg}$ using $g=9.8,\text{m/s}^2$.
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$49,\text{N}$. Weight calculation: $W=mg=5×9.8=49,\text{N}$.
$49,\text{N}$. Weight calculation: $W=mg=5×9.8=49,\text{N}$.
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What is the definition of a force vector’s magnitude and direction?
What is the definition of a force vector’s magnitude and direction?
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Magnitude is size in N; direction is the line and sense of push or pull. Magnitude measures strength; direction shows where force acts.
Magnitude is size in N; direction is the line and sense of push or pull. Magnitude measures strength; direction shows where force acts.
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What does it mean if the net force on an object is $0,\text{N}$?
What does it mean if the net force on an object is $0,\text{N}$?
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Forces balance; acceleration is $0,\text{m/s}^2$. No net force means no change in velocity per Newton's first law.
Forces balance; acceleration is $0,\text{m/s}^2$. No net force means no change in velocity per Newton's first law.
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State Newton’s second law in vector form relating net force and acceleration.
State Newton’s second law in vector form relating net force and acceleration.
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$\sum \vec{F}=m\vec{a}$. Vector form shows net force causes acceleration proportional to mass.
$\sum \vec{F}=m\vec{a}$. Vector form shows net force causes acceleration proportional to mass.
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What is the direction of an object’s acceleration compared with $\sum \vec{F}$?
What is the direction of an object’s acceleration compared with $\sum \vec{F}$?
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Acceleration is in the same direction as $\sum \vec{F}$. Newton's second law requires $\vec{a}$ parallel to $\sum \vec{F}$.
Acceleration is in the same direction as $\sum \vec{F}$. Newton's second law requires $\vec{a}$ parallel to $\sum \vec{F}$.
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Identify the net force when two collinear forces $10,\text{N}$ right and $6,\text{N}$ left act.
Identify the net force when two collinear forces $10,\text{N}$ right and $6,\text{N}$ left act.
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$4,\text{N}$ to the right. Subtract opposing forces: $10,\text{N} - 6,\text{N} = 4,\text{N}$ right.
$4,\text{N}$ to the right. Subtract opposing forces: $10,\text{N} - 6,\text{N} = 4,\text{N}$ right.
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Identify the net force when two collinear forces $8,\text{N}$ up and $8,\text{N}$ down act.
Identify the net force when two collinear forces $8,\text{N}$ up and $8,\text{N}$ down act.
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$0,\text{N}$. Equal opposite forces cancel: $8,\text{N} - 8,\text{N} = 0,\text{N}$.
$0,\text{N}$. Equal opposite forces cancel: $8,\text{N} - 8,\text{N} = 0,\text{N}$.
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State the action–reaction force rule (Newton’s third law) for a force pair.
State the action–reaction force rule (Newton’s third law) for a force pair.
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Equal magnitude, opposite direction, acting on different objects. Newton's third law: forces come in equal-opposite pairs on different bodies.
Equal magnitude, opposite direction, acting on different objects. Newton's third law: forces come in equal-opposite pairs on different bodies.
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Identify the direction of the normal force from a horizontal floor on a resting box.
Identify the direction of the normal force from a horizontal floor on a resting box.
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Upward, perpendicular to the surface. Normal force always pushes perpendicular away from surface.
Upward, perpendicular to the surface. Normal force always pushes perpendicular away from surface.
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State the magnitude of weight for mass $m$ near Earth’s surface.
State the magnitude of weight for mass $m$ near Earth’s surface.
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$F_g=mg$. Weight equals mass times gravitational acceleration.
$F_g=mg$. Weight equals mass times gravitational acceleration.
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Identify the weight of a $2.0,\text{kg}$ object if $g=9.8,\text{m/s}^2$.
Identify the weight of a $2.0,\text{kg}$ object if $g=9.8,\text{m/s}^2$.
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$19.6,\text{N}$ downward. Apply $F_g = mg$: $2.0,\text{kg} \times 9.8,\text{m/s}^2 = 19.6,\text{N}$.
$19.6,\text{N}$ downward. Apply $F_g = mg$: $2.0,\text{kg} \times 9.8,\text{m/s}^2 = 19.6,\text{N}$.
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