First slide

Static friction

Question

A force of 100 N is applied on a block of mass 3 kg as shown in the figure. The coefficient of friction between the surface and the block is $μ = \frac{1}{\sqrt{3}}$ . The frictional force acting on the block is

Moderate

A

15 N downwards
B

25 N upwards
C

20 N downwards
D

30 N upwards

Solution

$N^{'} = 100 \cos 30^{0} = 100 . \frac{\sqrt{3}}{2} = 50 \sqrt{3} N$
Net driving force = ( $100 \times \frac{1}{2} - 30$ )
= 20 N (upward)
$f_{limiting} \frac{1}{\sqrt{3}} (50 \sqrt{3}) = 50 N$
$∴ friction = 20 N (downwards)$

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