Search any question & find its solution
Question:
Answered & Verified by Expert
The minimum force required to move a body up an inclined plane is three times the minimum force required to prevent it from sliding down the plane. If the coefficient of friction between the body and the inclined plane is $\frac{1}{2 \sqrt{3}}$, then the angle of the inclined plane is
Options:
Solution:
1789 Upvotes
Verified Answer
The correct answer is:
$30^{\circ}$
Minimum force required to move a body up a rough inclined plane
$F_1=m g(\sin \theta+\mu \cos \theta)$
Minimum force required to prevent the body from sliding down the rough inclined plane
$F_2=\mu m g \cos \theta$
According to question,
$F_1=3 F_2$
$\begin{aligned} \therefore m g(\sin \theta+\mu \cos \theta) & =3(\mu m g \cos \theta) \\ \sin \theta+\mu \cos \theta & =3 \mu \cos \theta \\ \sin \theta & =2 \mu \cos \theta \\ \tan \theta & =2 \mu \\ & =2 \times \frac{1}{2 \sqrt{3}}=\frac{1}{\sqrt{3}} \\ & =\tan 30^{\circ} \\ \therefore \quad \theta & =30^{\circ}\end{aligned}$
$F_1=m g(\sin \theta+\mu \cos \theta)$
Minimum force required to prevent the body from sliding down the rough inclined plane
$F_2=\mu m g \cos \theta$
According to question,
$F_1=3 F_2$
$\begin{aligned} \therefore m g(\sin \theta+\mu \cos \theta) & =3(\mu m g \cos \theta) \\ \sin \theta+\mu \cos \theta & =3 \mu \cos \theta \\ \sin \theta & =2 \mu \cos \theta \\ \tan \theta & =2 \mu \\ & =2 \times \frac{1}{2 \sqrt{3}}=\frac{1}{\sqrt{3}} \\ & =\tan 30^{\circ} \\ \therefore \quad \theta & =30^{\circ}\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.