Forces, Rotation & Equilibrium — Chapter Practice, Grade 12

9 m/s²

a = ΣF ÷ m = 36 ÷ 4 = 9 m/s².

90 N

ΣF = ma = 0.045 × 2,000 = 90 N.

The book pulling up on the Earth

A third-law pair acts on two different objects. If Earth pulls the book down, the reaction is the book pulling Earth up. (The table's push is a separate force, not the reaction to gravity.)

Zero

Constant velocity means zero acceleration, so by ΣF = ma the net force must be zero — the cable tension exactly balances the weight.

Before the impact the passenger's head is at rest. When the car and seat are suddenly pushed forward, the head tends to stay where it was because of inertia (Newton's first law), so it lags behind the moving body and appears to snap backward. A headrest supplies the force needed to accelerate the head along with the body, preventing whiplash.

5 m/s²

Net force = 40 − 15 = 25 N east. a = ΣF ÷ m = 25 ÷ 5 = 5 m/s² east.

The reaction to a push acts on the other object, not automatically in this force sum.

F_net=ma=6 N in the acceleration direction.

State what the supplied observations or conditions establish, and separate that from any assumption.

Their interaction forces are equal and opposite; acceleration magnitude depends on each mass.

No. Different masses can have different accelerations under equal-magnitude forces, and other forces may also act.

Newton’s third law pairs the force B exerts on A with the force A exerts on B. Their magnitudes match and directions oppose, but they act on different bodies. They therefore do not cancel when calculating the acceleration of A alone. In a combined A+B system these mutual forces are internal and cancel in the total external-force account. Identify the agent and receiving body for every arrow before deciding which sum it belongs in.

No. Different masses can have different accelerations under equal-magnitude forces, and other forces may also act.

Response criteria

• Interpret labels, structures, axes, units and directions correctly.
• Explain the physical or environmental relationship or mechanism.
• Respect stated assumptions and distinguish observation from inference.

Net=12 N east; a=3 m/s² east. Its third-law partner is the cart’s 18 N force on the pulling agent, not the 6 N resistance on the cart.

Net=12 N east; a=3 m/s² east.

Its third-law partner is the cart’s 18 N force on the pulling agent, not the 6 N resistance on the cart.

Response criteria

• Use the stated measurements or conditions, with correct units.
• Connect the result to the physical or environmental mechanism or evidence relationship.
• Identify the assumption, model limit or further evidence requested.

30 N

Net force = 50 − 20 = 30 N to the right (the vertical weight and normal force cancel).

10 N

Perpendicular forces combine by the Pythagorean theorem: √(6² + 8²) = √(36 + 64) = √100 = 10 N.

Weight downward and the normal force upward

A resting book feels gravity pulling it down and the table's normal force pushing it up. The two are equal, so it stays in equilibrium.

add up to zero

Constant velocity means zero acceleration, so the net force is zero — the forces present must cancel out. (Forces can exist; they simply balance.)

Weight pulls straight down; the normal force from the snow pushes straight up; the rope's tension pulls forward (horizontal); and kinetic friction from the snow pushes backward, opposing the sled's motion. If the sled moves at steady speed, the up/down pair cancels and tension equals friction.

2 m/s²

Net force = √(12² + 5²) = √(144 + 25) = √169 = 13 N. Then a = ΣF ÷ m = 13 ÷ 6.5 = 2 m/s².

Normal equals weight only for the specified force and acceleration conditions.

30 N upward, balancing weight in the given model.

State what the supplied observations or conditions establish, and separate that from any assumption.

Newton's second law concerns that object's force sum; partner forces on other objects belong in other diagrams.

The upward tension component changes N, so using mg directly in μN would use the wrong contact force.

A vector equation becomes one scalar equation per perpendicular axis. Independent does not mean unrelated: the same tension angle enters both, and contact constraints can set a component of acceleration. For a block that remains on a horizontal table, ay=0 while ax may be nonzero. Assign signs to every component before substituting values; then solve the vertical support account before using a friction model that depends on N.

The upward tension component changes N, so using mg directly in μN would use the wrong contact force.

Response criteria

• Interpret labels, structures, axes, units and directions correctly.
• Explain the physical or environmental relationship or mechanism.
• Respect stated assumptions and distinguish observation from inference.

N=29.4−7.5=21.9 N. Horizontal net=15cos 30°−4≈8.99 N; ax≈3.00 m/s². The tension is counted once through its components.

N=29.4−7.5=21.9 N.

Horizontal net=15cos 30°−4≈8.99 N; ax≈3.00 m/s².

The tension is counted once through its components.

Response criteria

• Use the stated measurements or conditions, with correct units.
• Connect the result to the physical or environmental mechanism or evidence relationship.
• Identify the assumption, model limit or further evidence requested.

20 N

f = μₖN = 0.40 × 50 = 20 N.

36.75 N

N = mg = 15 × 9.8 = 147 N. f = μₖN = 0.25 × 147 = 36.75 N.

The maximum static friction is greater than the kinetic friction

Maximum static friction (μₛN) exceeds kinetic friction (μₖN), so it takes a bigger force to break the box free than to keep it sliding.

decreases

The normal force equals mg cosθ. As θ grows, cosθ shrinks, so the normal force decreases (reaching zero only at a vertical 90° 'ramp').

Static friction acts between surfaces that aren't sliding yet and adjusts up to a maximum to prevent motion; kinetic friction acts between surfaces already sliding and is a bit weaker. You feel the switch when pushing a heavy couch: it resists until you push past the static maximum, then suddenly 'gives' and slides more easily under the smaller kinetic friction.

3.2 m/s²

Along the slope: a = g(sinθ − μₖcosθ) = 9.8(0.5 − 0.20 × 0.866) = 9.8(0.5 − 0.173) = 9.8 × 0.327 ≈ 3.2 m/s². (Mass cancels out.)

Static friction instead adjusts up to its limiting value.

12 N, enough to balance the required tangential force without exceeding the maximum.

State what the supplied observations or conditions establish, and separate that from any assumption.

It changes the normal force and potentially the sliding tendency; friction depends on N and the motion regime.

Recalculate N using the vertical force balance; then update the kinetic-friction estimate and horizontal net force.

Kinetic friction acts during relative sliding. The school approximation fk≈μkN can model a limited range of surface conditions, although real friction may depend on speed, temperature, deformation and wear. On a level table with no vertical acceleration or extra vertical forces, N=mg here. That equality came from the force account, not from the friction law. State the sliding direction before assigning the sign of fk in Newton’s second law.

Recalculate N using the vertical force balance; then update the kinetic-friction estimate and horizontal net force.

Response criteria

• Interpret labels, structures, axes, units and directions correctly.
• Explain the physical or environmental relationship or mechanism.
• Respect stated assumptions and distinguish observation from inference.

N=49 N; static limit=19.6 N, so 15 N is balanced by 15 N friction. During sliding fk=14.7 N; a=(25−14.7)/5=2.06 m/s² along the push.

N=49 N; static limit=19.6 N, so 15 N is balanced by 15 N friction.

During sliding fk=14.7 N; a=(25−14.7)/5=2.06 m/s² along the push.

Response criteria

• Use the stated measurements or conditions, with correct units.
• Connect the result to the physical or environmental mechanism or evidence relationship.
• Identify the assumption, model limit or further evidence requested.

2 m/s²

a = v²/r = 10² ÷ 50 = 100 ÷ 50 = 2 m/s², directed toward the center of the curve.

18 N

F = mv²/r = 2 × 6² ÷ 4 = 2 × 36 ÷ 4 = 72 ÷ 4 = 18 N.

Toward the center of the circle

Uniform circular motion requires a center-seeking (centripetal) net force. The outward feeling is an illusion of the accelerating frame.

one-quarter as strong

Gravitation is an inverse-square law (F ∝ 1/r²). Doubling r multiplies the force by 1/2² = 1/4.

Even at constant speed the satellite's direction is always changing, and any change in velocity is acceleration — here it is centripetal acceleration pointing toward Earth's center. The force supplying it is gravity: Earth's gravitational pull provides exactly the mv²/r needed to bend the satellite's path into a circle, so it 'falls around' the planet.

6000 N

F = mv²/r = 1,200 × 20² ÷ 80 = 1,200 × 400 ÷ 80 = 1,200 × 5 = 6,000 N, directed toward the center of the curve.

Centripetal describes net inward force, not an additional mysterious interaction.

It quadruples force because F=mv²/r.

State what the supplied observations or conditions establish, and separate that from any assumption.

g=GM/r² decreases with distance from the center; doubling r gives one quarter the field.

r=2R, so g is one quarter of its surface value.

For ideal point masses or a spherical source outside its surface, Newtonian gravitation gives F=GMm/r² with center-to-center separation r. A satellite’s altitude h above a spherical Earth is not r; use r=R+h. Doubling center distance reduces force and gravitational acceleration to one quarter. The familiar near-surface g≈9.8 m/s² is an approximation over modest height differences, not the same value at every orbital distance.

r=2R, so g is one quarter of its surface value.

Response criteria

• Interpret labels, structures, axes, units and directions correctly.
• Explain the physical or environmental relationship or mechanism.
• Respect stated assumptions and distinguish observation from inference.

ac=16/2=8.0 m/s² and inward net force=0.50×8=4.0 N. Tension equals that net only if the specified geometry and other forces establish it as the sole inward contribution.

ac=16/2=8.0 m/s² and inward net force=0.50×8=4.0 N.

Tension equals that net only if the specified geometry and other forces establish it as the sole inward contribution.

Response criteria

• Use the stated measurements or conditions, with correct units.
• Connect the result to the physical or environmental mechanism or evidence relationship.
• Identify the assumption, model limit or further evidence requested.

Write the relationship first; substitute with consistent units.

3 N

100×.03=3.

The force lines can produce a nonzero net torque about the object.

Response criteria

• State a scientifically accurate conclusion.
• Show the calculation or mechanism that supports it.
• Include units, conditions, or limits where relevant.

List the given quantities, select a relationship, and check units before substituting.

1 J

½×200×.10²=1.

Force varies with extension; integrate or use the triangular graph area.

Response criteria

• State a scientifically accurate conclusion.
• Show the calculation or mechanism that supports it.
• Include units, conditions, or limits where relevant.

Use perpendicular lever arm and rotational sign.

F=kx=10 N, directed toward restoring equilibrium.

State what the supplied observations or conditions establish, and separate that from any assumption.

Zero net force prevents translational acceleration; zero net torque prevents angular acceleration.

No. The vector force sum must also vanish, and the starting state must be at rest.

Static equilibrium means the body remains at rest without translational or rotational acceleration. It needs ΣF=0 and Στ=0 about a selected pivot in the rigid-body model. Equal opposing torques alone can coexist with nonzero net force, while balanced forces can still form a turning couple. For this weightless-bar example equal downward loads at equal opposite arms balance torque, and the upward support must balance their combined 60 N force.

No. The vector force sum must also vanish, and the starting state must be at rest.

Response criteria

• Interpret labels, structures, axes, units and directions correctly.
• Explain the physical or environmental relationship or mechanism.
• Respect stated assumptions and distinguish observation from inference.

Torque balance: F×2=60×0.50, so F=15 N upward. Vertical balance gives pivot reaction=60−15=45 N upward. Both balances are necessary.

Torque balance: F×2=60×0.50, so F=15 N upward.

Vertical balance gives pivot reaction=60−15=45 N upward.

Both balances are necessary.

Response criteria

• Use the stated measurements or conditions, with correct units.
• Connect the result to the physical or environmental mechanism or evidence relationship.
• Identify the assumption, model limit or further evidence requested.

The rings push the loaded plate upward and Earth attracts that selected plate/load downward. Their magnitudes balance at rest but they are not a third-law pair. The plate pushes the rings downward as the support force’s partner on a different body. Impact peak force requires a time-resolved interaction and motion record, not a static layout.

Read source labels, scope, axes and intentional omissions.

Use the complete local explanation to draw a justified mechanism or calculate a supported quantity.

State what is observed, what is assumed and what evidence remains needed.

Response criteria

• Interpret the source labels, units and scope accurately.
• Explain the relevant mechanism with a labeled model or supported calculation.
• Identify a meaningful assumption, uncertainty or missing comparison.