First slide

Friction on inclined plane of angle more than angle of repose

Question

A 30 kg box has to move up an inclined slope of $30^{0}$ to the horizontal at a uniform velocity of 5 $m s^{- 1}$ . If the frictional force retarding the motion is 150N, the horizontal force required to move up is (g=10 $m s^{- 2}$ )

Moderate

A
$300 \times \frac{2}{\sqrt{3}} N$
B
$300 \times \frac{\sqrt{3}}{2} N$
C

300N
D

150N

Solution

Fcos300 = mgsinθ + f
Fcos300 = mgsin $30^{0}$ + f
$F \frac{\sqrt{3}}{2} = 30 \times 10 \times \frac{1}{2} + 150 = 300 \Rightarrow F = 300 \times \frac{2}{\sqrt{3}} N$

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