2.4 Newton’s First Law — Worked Solutions
Work through Topic 2.4
These are the full worked solutions to the lesson questions. Try the questions first, then check each step and the reason for it.
Question 1
An object moves at constant velocity in an inertial frame. What is its net force?
- Zero
- In the direction of motion
- Equal to its weight in all cases
Step 1. Constant velocity means both its speed and direction stay unchanged, so a = 0.
Why this step? The question specifies velocity, not merely constant speed along a curved path.
Step 2. In the inertial frame, ΣF = ma = 0.
Why this step? Newton’s laws apply in this frame; individual nonzero forces may cancel even though their resultant vanishes.
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Question 2
A ball is momentarily at rest at the top of a vertical toss. Is it in translational equilibrium?
- Yes, because velocity is zero
- No, because gravity gives a nonzero net force
- Yes, because acceleration is temporarily zero
Step 1. At the toss peak, velocity is zero for an instant, but Earth still exerts mg downward.
Why this step? The momentary turning condition supplies no upward contact or other balancing force.
Step 2. The net force is therefore nonzero and a = −g with up positive.
Why this step? Translational equilibrium is a zero-force condition; it cannot be inferred from one instantaneous zero-velocity measurement.
Step 3. The ball is not in translational equilibrium.
Why this step? Its velocity changes immediately after the peak, which is exactly the behavior the unbalanced gravitational force predicts.
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Question 3
Vertical forces on a cart balance, but a horizontal pull is unbalanced. What can you conclude?
- All velocity components stay constant
- Only the vertical velocity component remains unchanged
- The cart cannot accelerate horizontally
Step 1. Vertical balance gives ΣF_y = 0, so a_y = 0 and v_y remains unchanged.
Why this step? The signed vertical force sum controls only the vertical component; it does not settle forces in the perpendicular axis.
Step 2. The unbalanced horizontal pull gives a_x ≠ 0 and changes v_x.
Why this step? The separate horizontal second-law equation has a nonzero right-hand side, so the full velocity vector is not constant.
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Question 4
A 4 kg lamp hangs from one vertical cable. Find tension. Describe how the FBD differs if the lamp moves upward at a constant velocity.
Step 1. For the stationary lamp, T − mg = 0, giving T = 4·10 = 40 N.
Why this step? The one vertical cable supplies the only stated upward force, and zero acceleration requires it to match weight.
Step 2. For constant upward velocity, a is still zero, so the same two-arrow FBD and T = 40 N apply.
Why this step? Newton’s law depends on velocity change rather than on whether the lamp is already moving upward; constant motion adds no force of motion.
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Question 5
A cart has forces 5 N east and 5 N west, plus 8 N upward and 8 N downward. Can you determine its velocity from those forces? Explain.
Step 1. Add east-west components: 5 − 5 = 0 N. Add vertical components: 8 − 8 = 0 N.
Why this step? The opposing forces cancel within each axis, so the full vector resultant is zero rather than just one component.
Step 2. The cart has zero acceleration, so its velocity stays at its initial value.
Why this step? In an inertial frame, zero resultant stops the velocity from changing but does not reset that value to zero.
Step 3. You cannot determine that initial velocity from the given forces.
Why this step? A resting cart and a steadily moving cart can have the same balanced force configuration; a position-time or initial-velocity measurement is still needed.
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Question 6
Derive the cable tension for a stationary sign of mass m suspended by two symmetric cables at angle θ above horizontal. Predict the factor of change when m doubles at fixed angle.
Step 1. By symmetry, both cables have tension T. Their horizontal components are equal and opposite.
Why this step? The symmetric stationary arrangement cannot have an uncanceled sideways resultant.
Step 2. Vertical balance is T sin θ + T sin θ − mg = 0.
Why this step? Angles are measured above horizontal, so sine selects each upward component; both cables share the supporting load.
Step 3. Solve T = mg/(2 sin θ), for a supporting angle with sin θ > 0.
Why this step? Dividing by the nonzero upward-component factor isolates the full cable force; a horizontal cable has no finite supporting component.
Step 4. At fixed θ, doubling m changes T to 2mg/(2 sin θ) = 2T.
Why this step? Only the weight factor changes, while the two-cable geometry stays fixed, making tension directly proportional to mass.