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=2 F _2$
$ \frac{ mg }{\sqrt{2}}(1+\mu)=2 \frac{ mg }{\sqrt{2}}(1-\mu) $
$ 1+\mu=2-2 \mu $
$ \mu=1 / 3=0.33$