two trunks sit side by side on the floor. the larger trunk (52 kg) is to the left of the smaller trunk (34 kg). a person pushes on the larger trunk horizontally toward the right. the coefficient of static friction between the trunks and the floor is 0.35. (a) determine the magnitude of the maximum force the person can exert without moving either trunk. ans: 290 n (b) calculate the force the larger trunk exerts on the smaller trunk. ans: 120 n right (c) would either answer change if the person pushed in the opposite direction on the smaller trunk? explain your reasoning

Question

Accepted Answer

### Part (a)
# Explanation:
## Step1: Identify total normal force
The normal force on each trunk is equal to its weight. Total mass $ m_{total} = 52\,\text{kg} + 34\,\text{kg} = 86\,\text{kg} $. Total normal force $ F_N = m_{total}g $, where $ g = 9.8\,\text{m/s}^2 $. So $ F_N = 86 \times 9.8 = 842.8\,\text{N} $.
## Step2: Calculate maximum static friction
Maximum static friction $ F_{f,\text{max}} = \mu_s F_N $, with $ \mu_s = 0.35 $. Thus $ F_{f,\text{max}} = 0.35 \times 842.8 \approx 295\,\text{N} $ (close to 290 N, likely due to g approximation). The maximum force without moving is equal to this friction.
# Answer: $\approx 290\,\text{N}$ (or more precisely ~295 N, but as per given ans: 290 N)

### Part (b)
# Explanation:
## Step1: Analyze smaller trunk
For the smaller trunk (34 kg), the force from the larger trunk ($ F $) must overcome its static friction. Normal force on smaller trunk $ F_{N,small} = 34 \times 9.8 = 333.2\,\text{N} $. Maximum static friction on smaller trunk $ F_{f,small} = 0.35 \times 333.2 \approx 116.62\,\text{N} \approx 120\,\text{N} $. This is the force the larger trunk exerts (since it's the force needed to just not move the smaller trunk, equal to its static friction).
# Answer: $ 120\,\text{N} $ [right]

### Part (c)
# Explanation:
## Step1: Analyze total friction (for part a)
If pushing the smaller trunk left, total mass and normal force remain same, so total static friction (max force) remains $ \mu_s (m_1 + m_2)g $, so part (a) answer doesn't change.
## Step2: Analyze force on larger trunk (for part b)
Now, the force on the larger trunk would be due to its static friction. Larger trunk's normal force $ F_{N,large} = 52 \times 9.8 = 509.6\,\text{N} $, its static friction $ F_{f,large} = 0.35 \times 509.6 \approx 178.36\,\text{N} $, but wait—no, when pushing smaller trunk left, the force between trunks would be limited by the smaller trunk's friction? Wait, no: when pushing smaller trunk left, the larger trunk's friction opposes, and the force between them would be based on the smaller trunk's friction? Wait, actually, total maximum force still depends on total mass (same total normal force), so part (a) max force (total friction) remains. For part (b), now the force the smaller trunk exerts on the larger trunk would be limited by the larger trunk's static friction? Wait, no—let's re-express:

When pushing smaller trunk left, the system's total friction is still total mass × g × μ_s (same as before, since total mass doesn't change). For the force between trunks: now, the larger trunk's static friction is $ \mu_s m_{large}g = 0.35×52×9.8≈179\,\text{N} $, and the smaller trunk's friction is $ \mu_s m_{small}g≈117\,\text{N} $. The force between them would be limited by the smaller trunk's friction? No—wait, if pushing smaller trunk left, the larger trunk is to the left of the smaller? Wait, original setup: larger (52 kg) left of smaller (34 kg). So pushing smaller trunk left: the smaller trunk would push the larger trunk left? No, wait, direction: original push is right on larger. If push left on smaller, the smaller trunk would exert a force on the larger trunk. But the total maximum force (part a) is still total static friction (same total mass). For the force between trunks: now, the force from smaller to larger would be limited by the larger trunk's static friction? Wait, no—let's think of free body diagrams.

If pushing smaller trunk left (toward larger trunk), the force between them: the larger trunk's static friction (opposing leftward force) is $ F_{f,large} = \mu_s m_{large}g = 0.35×52×9.8≈179\,\text{N} $, and the smaller trunk's static friction (opposing leftward push) is $ F_{f,small} = \mu_s m_{small}g≈117\,\text{N} $. The maximum force before moving: total friction is still total mass × g × μ_s (same as before, since both trunks' frictions add up to total friction). But the force between trunks: when pushing smaller left, the force the smaller exerts on the larger is limited by the smaller's static friction? No, actually, the force between them must balance the friction of the trunk being pushed against. Wait, maybe simpler:

- **Part (a)**: Total maximum force (to not move either) depends on total static friction, which is $ \mu_s (m_1 + m_2)g $. Since total mass is same, this doesn't change. So part (a) answer doesn't change.
- **Part (b)**: Now, the force between trunks is limited by the larger trunk's static friction (since we're pushing the smaller trunk left, the larger trunk's friction opposes the force from the smaller trunk). Larger trunk's static friction: $ F_{f,large} = \mu_s m_{large}g = 0.35×52×9.8≈179\,\text{N} $, but wait, no—wait, when pushing smaller left, the smaller trunk is trying to move left, so the larger trunk exerts a force on the smaller trunk? No, direction: larger is left of smaller. Pushing smaller left (toward larger), so smaller trunk pushes larger trunk left. The larger trunk's static friction (opposing left) is its maximum friction. But the force between them would be limited by the smaller trunk's ability to push, which is limited by its own static friction? Wait, no—let's use the same logic as before, but reversed. The force between trunks is now limited by the larger trunk's static friction? Wait, no, the key is: when pushing the smaller trunk left, the system's total friction is still total mass × μ_s × g (same as before, because total normal force is same). For the force between trunks: originally, it was the smaller trunk's friction; now, it's the larger trunk's friction? Wait, no—let's recast.

Original (pushing larger right): force between trunks is to overcome smaller's friction.
New (pushing smaller left): force between trunks is to overcome larger's friction. Larger trunk's friction: $ \mu_s m_{large}g = 0.35×52×9.8≈179\,\text{N} $, but the given answer for part (b) was 120 N (smaller's friction). So now, the force between trunks would be larger (≈180 N), so part (b) answer changes. Wait, but let's check:

Wait, when pushing the smaller trunk left, the two trunks are still side by side. The maximum force without moving either trunk: total friction is still total mass × μ_s × g (same as part a, so part a answer doesn't change). But the force between trunks: now, the force the smaller trunk exerts on the larger trunk must be equal to the larger trunk's static friction (since the larger trunk's friction opposes the leftward push from the smaller trunk). So larger trunk's static friction: $ F_{f,large} = 0.35×52×9.8≈179\,\text{N} $, but the original part (b) was 120 N (smaller's friction). So part (b) answer would change, while part (a) remains the same.

Wait, but let's confirm:

- **Part (a)**: Total maximum force is total static friction, which depends on total mass (52+34) and μ_s. So regardless of which trunk is pushed, total mass is same, so total normal force is same, so total static friction (max force) is same. So part (a) answer does not change.
- **Part (b)**: When pushing larger right, the force between trunks is limited by the smaller trunk's static friction (since the larger trunk pushes the smaller, so the smaller's friction resists). When pushing smaller left, the force between trunks is limited by the larger trunk's static friction (since the smaller trunk pushes the larger, so the larger's friction resists). Larger trunk's static friction: $ \mu_s m_{large}g = 0.35×52×9.8≈179\,\text{N} $, which is different from the smaller's 120 N. So part (b) answer changes.

So explanation:
- For part (a): Total maximum force depends on total mass (to find total normal force) and static friction coefficient. Since total mass ($ 52 + 34 = 86\,\text{kg} $) and $ \mu_s $ don’t change, total static friction ($ F_{f,\text{max}} = \mu_s (m_1 + m_2)g $) remains the same. Thus, the maximum force without moving either trunk (part a) does not change.
- For part (b): The force between trunks is limited by the static friction of the trunk being "pushed against" (originally the smaller trunk, now the larger trunk). The larger trunk has more mass, so its static friction ($ \mu_s m_{\text{large}}g $) is greater than the smaller trunk’s ($ \mu_s m_{\text{small}}g $). Thus, the force between trunks (part b) changes (from ~120 N to ~180 N).

# Answer (Part c):
- Part (a) answer **does not change** (total mass and normal force remain same, so total static friction is same).
- Part (b) answer **changes** (force between trunks is now limited by the larger trunk’s static friction, which is greater than the smaller trunk’s, so the force magnitude increases).

Explanation: Total maximum force (part a) depends on total mass (same for both push directions), so it’s unchanged. The force between trunks (part b) depends on the static friction of the trunk receiving the internal force: when pushing the smaller trunk left, the larger trunk’s static friction (larger mass) determines the force, which is different from the smaller trunk’s friction (when pushing larger right).

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QUESTION IMAGE

Question

Explanation:

Part (a)

Step1: Identify total normal force

Step2: Calculate maximum static friction

Step1: Analyze smaller trunk

Step1: Analyze total friction (for part a)

Step2: Analyze force on larger trunk (for part b)

Answer:

Part (b)