Question

a. Block A as shown in figure, weights 60.0N. The coefficient of static friction between the block and the surface on which it rests is 0.25. The weight w is 12.0 N and the system is in equilibrium. Find the friction force exerted on block A.
b. Find the maximum weight w for which the system will remain in equilibrium.

Answer 1

(a) The tension in wire T attached to the hook can be determined as,

$Tsin\left(\theta \right)w$

$T=\frac{w}{sin\left(\theta \right)}$

The force due to static friction on block A is equal to the horizontal component of the

tension.

The expression for frictional force is,

$F_f=Tcos\left(\theta \right)$

Substitute $T=\frac{w}{sin\left(\theta \right)}$ in the frictional force expression

$F_f=\frac{w}{sin\left(\theta \right)}cos\left(\theta \right)$

$=wcot\left(\theta \right)$

Substitute 12 N for w and ${45}^{\circ }$ for $\theta .$

$F_f=\left(12N\right)cot\left({45}^{\circ }\right)$

$=12N$

Therefore friction force exerted on block A is 12N.

(b)

From the system to remain in equilibrium force of friction should be eqy=ual to the weight w of the block.

The frictional forceis,

$F_f=m_{A}g{\mu }_2$

Here, $m_A$ is mass of A, g is acceleration due to gravity and ${\mu }_s$ is coefficient

of static friction.

Substitute 60 N for mg and 0.25 for ${\mu }_s.$

$F_f=\left(60N\right)\left(0.25\right)$

$=15N$

The force of friction should be equal to the weight w of the block so the weight of the

block is also 15 N.

Therefore, maximum weight w for the system to remain in equilibrium is 15 N.