First slide

Friction on horizontal surface

Question

In the figure given, the system is in equilibrium. What is the maximum value that W can have if the friction force on the 40 N block cannot exceed 12.0 N

Moderate

A

3.45 N
B

6.92 N
C

10.35 N
D

12.32 N

Solution

The free-body diagram of the block is as shown in adjacent figure.

$\begin{array}{l} ∵ {(f_{0})}_{max} = 12 .0 N (Given) \\ ∵ {(T_{1})}_{max} = 12 .0 N \end{array}$
The free-body diagram of knot is as shown in adjacent figure.

$\begin{aligned} T_{2} \sin 30^{\circ} = W \dots (i) \\ T_{2} \cos 30^{\circ} = T_{1} \dots (ii) \end{aligned}$
Divide (i) by (ii), we get, $\tan 30^{\circ} = \frac{W}{T_{1}}$
or $W = T_{1} \tan 30^{\circ} = 0 .577 T_{1}$
$∴ W_{max} = 0 .577 {(T_{1})}_{max} = 0 .577 \times 12 .0 N = 6 .92 N$

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