2.2 Forces and Free-Body Diagrams

1. AEarth's pull, the ice's push and a fading push left over from the hit
   A student who thinks a hit gives the puck a force that it carries along picks this. The stick exerts a force only while it touches the puck. The puck keeps sliding because nothing slows it, not because a force pushes it forward.
2. BA downward pull by Earth and an upward push by the ice that touches it Correct
   Each force needs an object that exerts it. After the hit only two objects interact with the puck: Earth, which pulls it down, and the ice, which touches it and pushes it up. The stick no longer touches the puck, so it exerts no force on it.
3. CA downward pull by Earth, and no force at all from the ice
   A student who thinks a surface only blocks and cannot push picks this. The ice is in contact with the puck and pushes up on it; this upward contact force is why the puck does not sink into the ice.
4. DAn upward push by the ice, with no gravitational force on a supported puck
   A student who thinks gravity acts only on falling objects picks this. Earth pulls on the puck whether or not it is falling; the ice's push acts alongside Earth's pull, it does not replace it.

CED 2.2.A.1.i · Read this in Fix

1. AThe table pushes up on the book by means of electric forces between their atoms. Correct
   The table and the book touch, so the table exerts a contact force on the book. Contact forces are the large-scale result of electric forces between atoms: the book presses the atoms of the tabletop slightly closer together, and the electric forces between the atoms of the two surfaces push the book up.
2. BThe table pushes up on the book with Earth's gravity, sent back up through it.
   A student who thinks a surface's push is gravity passed back up picks this. Gravity is an interaction between the book and Earth; the table's push is a separate interaction, a contact force between the table and the book that is electric in origin.
3. CThe table exerts no force, since a rigid object only blocks the book's way.
   A student who thinks a rigid support cannot push picks this. No real table is perfectly rigid: its surface is compressed very slightly, and the electric forces between atoms push back on the book. That upward push is what stops the book from falling through the table.
4. DThe table pushed on the book as it landed, but exerts no force now it is at rest.
   A student who thinks forces act only during events such as landing picks this. The table pushes up on the book for as long as the book rests on it; without that push, the book would fall through.

CED 2.2.A.2 · Read this in Fix

1. A25 N Correct
   Only F_T and F_f have horizontal components. The horizontal component of F_T is the side adjacent to the 37° angle, 50 N × cos 37° = 40 N to the right. Friction contributes 15 N to the left, so the sum is 40 N − 15 N = 25 N, directed to the right.
2. B15 N
   A student who uses sin 37° for the horizontal component picks this: 50 N × 0.60 − 15 N = 15 N. The angle is measured from the horizontal, so the horizontal component is the adjacent side, F_T cos 37° = 40 N; F_T sin 37° = 30 N is the vertical component.
3. C55 N
   A student who adds magnitudes whatever their directions picks this: 40 N + 15 N = 55 N. Friction points to the left, opposite the rope's horizontal component, so it subtracts: 40 N − 15 N = 25 N.
4. D65 N
   A student who counts the rope's horizontal component as an extra force, alongside the rope's force itself, picks this: 40 N + 40 N − 15 N = 65 N. The component is part of F_T, not a second force, so it is counted once.

Working Take +x to the right. F_T: F_T cos 37° = 50 N × 0.80 = +40 N. F_f: −15 N. F_g and F_N are vertical, so their horizontal components are zero. Sum = 40 N − 15 N = +25 N, so the magnitude is 25 N (directed to the right).

CED 2.2.B.1 · Read this in Fix

1. ADiagram 1
   A student who thinks gravity does not act on a supported object picks the diagram with F_T alone. Earth pulls on the crate even while the rope holds it, so F_g must be drawn.
2. BDiagram 2
   A student who includes forces exerted by the crate picks the diagram with three arrows. F_crate on rope is exerted on the rope, so it belongs on the rope's diagram; the crate's diagram shows only forces exerted on the crate.
3. CDiagram 3 Correct
   The crate interacts with two objects: Earth, which pulls it down (F_g), and the rope, which touches it and pulls it up (F_T). The correct diagram shows exactly these two forces as arrows from the dot, of equal length because the crate stays at rest.
4. DDiagram 4
   A student who thinks a rope only holds and cannot pull picks the diagram with F_g alone. The rope is in contact with the crate and pulls up on it with the force F_T, which must be drawn.

Working The crate interacts with two objects: Earth (gravitational force F_g, downward) and the rope (tension F_T, upward); nothing else touches it. F_crate on rope is exerted on the rope, not on the crate. The correct diagram therefore has exactly two arrows from the dot, F_T up and F_g down, of equal length because the crate stays at rest.

CED 2.2.B.2 · Read this in Fix

1. AKeep all three arrows, as the rope pulls the crate both forward and upward.
   A student who treats components as extra forces picks this. The rope's forward and upward effects are both contained in the single force F_T; drawing the components as well would count the rope's pull twice.
2. BRemove the two component arrows; they are parts of F_T, not extra forces. Correct
   The rope exerts one force on the crate, F_T. Its horizontal and vertical components describe that same force along two axes, so drawing them as well shows the rope's pull twice. On an AP free-body diagram each force is a single arrow from the dot; components are used in the equations, not drawn.
3. CRemove F_T and keep its components, as forces must be horizontal or vertical.
   A student who thinks forces must be horizontal or vertical picks this. A free-body diagram shows each force in its actual direction, so the rope's force is drawn along the rope, at its angle.
4. DReplace all three with a single arrow for the net force on the crate.
   A student who thinks the net force belongs on the diagram picks this. A free-body diagram shows each force exerted by the environment; the net force is found by adding those forces, not drawn in place of them.

CED 2.2.B.3 · Read this in Fix

1. AA's result is smaller, as tilting the axes shrinks the gravitational force.
   A student who thinks the choice of axes changes the forces picks this. Tilting the axes changes how the gravitational force is split into components, not the force itself, so both students work with the same forces and find the same normal force.
2. BThe results are equal, but A's equations are simpler to solve. Correct
   The normal force is a physical quantity, and choosing axes cannot change it, so correct work with either set of axes gives the same magnitude. A's axes put the block's acceleration along one axis, so A's perpendicular equation contains the normal force alone; B must solve two equations that each contain both the normal force and the acceleration.
3. CB's result is the valid one, since axes must be horizontal and vertical.
   A student who thinks axes must be horizontal and vertical picks this. Any pair of perpendicular axes is valid; tilted axes are simply more convenient here, because the block accelerates along the ramp.
4. DB's result is larger, since the normal force's two components add up to more.
   A student who adds components as if they were plain numbers picks this. Components combine as the sides of a right triangle, not by adding magnitudes, so splitting the normal force into horizontal and vertical parts does not change its magnitude.

CED 2.2.B.4 · Read this in Fix

1. AThe force on cart 1 is larger than the force on cart 2.
   A student who reads a negative component as a smaller force picks this. The sign is a direction, not a size: both forces have a magnitude of 6 N.
2. BThey are the same force, since each has a size of 6 N.
   A student who compares forces by their sizes alone picks this. A force is a vector: two forces of equal magnitude that point in opposite directions are different forces, with opposite effects.
3. CBoth are 6 N in magnitude but opposite in direction. Correct
   In one dimension the sign of a force component gives its direction relative to the chosen axis. Both forces have a magnitude of 6 N; the force on cart 1 points to the right (+x) and the force on cart 2 points to the left (−x).
4. DCart 2 must be slowing down while cart 1 speeds up.
   A student who links a negative force with slowing down picks this. The sign gives only the force's direction along the axis. Whether a cart slows down or speeds up depends on the net force on it compared with its direction of motion, and the stem gives neither; a cart moving to the left with a net force to the left speeds up.

Working In one dimension the sign of a component gives the direction along the axis and the number without its sign gives the magnitude: |+6 N| = |−6 N| = 6 N; +6 N points in +x (right), −6 N points in −x (left).

CED 2.2.A.1 · Read this in Fix

1. ARemove F_kick, since no object exerts it once the foot has left the ball. Correct
   A force is always due to an interaction with another object. The foot interacted with the ball only while they touched; now the only object interacting with the ball is Earth. The correct diagram has a single arrow, F_g, pointing down.
2. BKeep both, since a ball moving up and to the right needs a force that way.
   A student who thinks motion needs a force in its direction picks this. The ball keeps moving up and to the right because it already has that velocity; no object pushes it along its path, so there is no force in that direction.
3. CRemove F_g, since Earth pulls on the ball only once it starts to fall.
   A student who thinks gravity acts only on falling objects picks this. Earth pulls the ball down throughout its flight, while it rises as well as while it falls; that pull is what slows its rise.
4. DAdd an upward arrow for the gravitational force that the ball exerts on Earth.
   A student who includes forces exerted by the object picks this. The ball's pull on Earth is exerted on Earth, so it belongs on Earth's free-body diagram, not on the ball's.

CED 2.2.A.1.i · Read this in Fix

1. AYes, because the push is exerted on the spacecraft, which is part of the system.
   A student who thinks a push on any part of a system acts on the whole system picks this. The push does act on the spacecraft, but it is exerted by the astronaut, who is also inside the system; the wall pushes back on the astronaut, and forces between parts of a system give no net force on the system.
2. BYes, because the astronaut pushes on the wall harder than it pushes back.
   A student who thinks the active pusher exerts the larger force picks this. Both forces are exerted between parts of the system, so they are internal and cannot give the system a net force; in fact they also have equal magnitudes, as topic 2.3 shows.
3. CNo, because with no air or gravity around, a push has nothing to act on.
   A student who thinks a push needs air or gravity picks this. A push needs only the two objects involved: in empty space the astronaut and the spacecraft really are pushed apart. There is no net force on the system for a different reason: both forces are internal.
4. DNo, because the push is an interaction between two parts of the system. Correct
   The astronaut and the wall are both inside the system, so the astronaut's push on the wall and the wall's push on the astronaut are internal forces. An object or system cannot exert a net force on itself: the astronaut and the spacecraft are pushed apart, but the system as a whole receives no net force.

CED 2.2.A.1.ii · Read this in Fix

1. AIt is missing F_P on Q, the downward push exerted on Q by P.
   A student who draws the forces that objects inside the system exert on each other picks this. P's push on Q, like Q's push on P, is exerted by one part of the system P + Q on another part, so it is internal and is left off the system's diagram; only forces from outside the system (Earth, the floor and the hand) appear.
2. BF_N should start at the bottom of Q, where the floor touches it.
   A student who thinks each force must start where it is applied picks this. The system is treated as though all its mass is at its center of mass, shown as a dot, and every force, including the floor's push, is drawn from the dot.
3. CIt needs one more arrow, for the net force that makes the system speed up.
   A student who draws the net force as a force of its own picks this. The net force is the sum of the forces already shown, not a force exerted by some object; drawing it as well would misrepresent the forces on the system.
4. DRemove F_Q on P, which one part of the system exerts on another part. Correct
   Q's upward push on P is exerted by one part of the system on another part, so it is an internal force. A system's free-body diagram shows only forces exerted by objects outside it: here Earth's pull, the floor's push and the hand's push.

CED 2.2.B.3 · Read this in Fix

1. A+F_g cos θ
   A student who uses cos θ for an x-component out of habit picks this. Here θ is not measured from the x-axis: F_g cos θ is the size of the component perpendicular to the ramp, and it would wrongly give the whole of F_g along a level surface (θ = 0).
2. B+F_g tan θ
   A student who treats F_g as one of the perpendicular sides of the triangle, the side adjacent to θ, picks this. F_g is the hypotenuse of the triangle it forms with its components, so the component along the ramp is F_g sin θ; F_g tan θ would be larger than F_g on a ramp steeper than 45°, which no component can be.
3. C+F_g sin θ Correct
   The angle between F_g and the perpendicular to the ramp equals the ramp angle θ, so the component along the ramp is the side of the triangle opposite θ, F_g sin θ. It points down the ramp, the +x direction, so it is positive. On a level surface (θ = 0) it correctly becomes zero.
4. D+F_g/sin θ
   A student who divides by the trigonometric ratio picks this. F_g is the hypotenuse of the triangle it forms with its components, so each component is F_g multiplied by sin θ or cos θ; F_g/sin θ is larger than F_g, which no component can be.

Working The +y axis is perpendicular to the ramp, so the angle between F_g (straight down) and the −y direction equals the ramp angle θ. F_g is the hypotenuse of the right triangle it forms with its components: the component along −y (adjacent to θ) has magnitude F_g cos θ, and the component along the ramp (opposite θ) has magnitude F_g sin θ. F_g points partly down the ramp, the +x direction, so F_gx = +F_g sin θ. Check: on a level surface, θ = 0, the component along the surface is F_g sin 0 = 0, as it must be.

CED 2.2.B.4 · Read this in Fix

1. AEarth's pull on B, the floor's upward push on B, and Earth's pull on A
   A student who thinks A's weight is exerted on B picks this. Earth's pull on A is exerted on A. What acts on B is a different force, A's contact push on B, which here has the same size as A's weight because A is at rest.
2. BEarth's pull on B and the floor's push on B, as a resting A exerts no force
   A student who thinks a resting object exerts no force picks this. A presses on B's top surface, so B's diagram must include A's downward push, just as it includes the floor's push.
3. CEarth's pull on B, the floor's upward push on B, and B's push on the floor
   A student who includes forces exerted by B picks this. B's push on the floor is exerted on the floor, so it belongs on the floor's diagram, not on B's.
4. DEarth's pull on B, the floor's upward push on B, and the downward push of A on B Correct
   B interacts with three objects: Earth (gravitational force), the floor (upward contact force) and block A, which touches B and pushes down on it. All three forces are exerted on B, so all three belong on B's diagram.

CED 2.2.B.2 · Read this in Fix