Question

Consider a block kept on an inclined plane (inclined at $45^{\circ }$ ) as shown in the figure. If the force required to just push it up the incline is $2$ times the force required to just prevent it from sliding down, the coefficient of friction between the block and inclined plane $(\mu )$ is equal to :

• A. $0.50$
• B. $0.33$
• C. $0.25$
• D. $0.60$

Answer 1

Answer: (B)

$F_1=mg\sin 45^{\circ }+f=mg\sin 45^{\circ }+\mu N$
$F_1=\frac{mg}{\sqrt{2}}+\mu mg\cos 45^{\circ }$
$F_1=\frac{mg}{\sqrt{2}}(1+\mu )$

$F_2=mg\sin 45^{\circ }-f=mg\sin 45^{\circ }-\mu N$
$=\frac{mg}{\sqrt{2}}(1-\mu )$
$F_1=2F_2$
$\frac{mg}{\sqrt{2}}(1+\mu )=2\frac{mg}{\sqrt{2}}(1-\mu )$
$1+\mu =2-2\mu$
$\mu =1/3=0.33$