Question

A force of 525 N is applied to a stationary couch of mass 100 kg, where μs = 0.40 and μk = 0.32. What is the acceleration of the couch

Answer 1

Answer:

Approximately $2.11\; \rm m\cdot s^{-2}$ (assuming that $g = 9.81\; \rm m\cdot s^{-2}$, the floor is horizontal, and that the external force is applied horizontally.)

Explanation:

The mass of this couch is $m = 100\; \rm kg$. Calculate the weight of the couch:

$W = m \cdot g = 100\; \rm kg \times 9.81\; \rm m \cdot s^{-2} = 981\; \rm N$.

Assume that the floor is horizontal. The magnitude of the normal force between the floor and the couch would be equal to the size of the weight of this couch.

Therefore, the normal force between the floor and the couch would be $N = 981\; \rm N$.

The first step is to find out whether the couch would move at all.

The question has provided two constants of friction: $\mu_\text{s}$ (coefficient of static friction) and $\mu_\text{k}$ (coefficient of kinetic friction.)

The coefficient of static friction applies when the couch is stationary relative to the floor: $f_\text{s} = \mu_\text{s} \cdot N = 0.40\times 981\; \rm N \approx 313.92\; \rm N$. In other words, while the couch isn't moving, the maximum (horizontal) external force that friction could resist would be $f_\text{s} = 313.92\; \rm N$.

Assume that the external force in this question is horizontal. The size of the external force ($525\; \rm N$) exceeds that of $f_\text{s}$. Hence, the couch would start to move after the $525\; \rm N\!$ external force is applied.

Once the couch starts to move, the coefficient of static friction would no longer be relevant. Instead, the coefficient of kinetic friction ($\mu_\text{k}$) would give size of the friction between the floor and the couch. The size of that friction would be $f_\text{k} = \mu_\text{k} \cdot N = 0.32\times 981\; \rm N \approx 313.92\; \rm N$.

This friction would counteract the horizontal external force on the couch. Hence, the net force on the couch in the horizontal direction would be:

$\begin{aligned}& F(\text{external}) - f_\text{k} \\ &\approx 525\; \rm N - 313.92\; \rm N \\ &\approx 211.08\; \rm N \end{aligned}$.

Apply Newton's Second Law of motion to find the acceleration of the couch:

$\begin{aligned}a &= \frac{(\text{Net Force})}{m} \\ & \approx \frac{211.08\; \rm N}{100\; \rm kg} \approx 2.11\; \rm m \cdot s^{-2}\end{aligned}$.