Search any question & find its solution
Question:
Answered & Verified by Expert
A rectangular box lies on a rough inclined surface. The coefficient of friction between the surface and the box is $\mu$. Let the mass of the box be $m$.
(a) At what angle of inclination $\theta$ of the plane to the horizontal will the box just start to slide down the plane?
(b) What is the force acting on the box down the plane, if the angle of inclination of the plane is increased to $\alpha>\theta$ ?
(c) What is the force needed to be applied towards along the plane to make the box either remain stationary or just move up with uniform speed?
(d) What is the force needed to be applied upwards along the plane to make the box move up the plane with acceleration $a$ ?
(a) At what angle of inclination $\theta$ of the plane to the horizontal will the box just start to slide down the plane?
(b) What is the force acting on the box down the plane, if the angle of inclination of the plane is increased to $\alpha>\theta$ ?
(c) What is the force needed to be applied towards along the plane to make the box either remain stationary or just move up with uniform speed?
(d) What is the force needed to be applied upwards along the plane to make the box move up the plane with acceleration $a$ ?
Solution:
1773 Upvotes
Verified Answer
(a) Force of friction on the box will act up the plane.
For the box to just starts sliding down $(m g)$ the plane then
$$
\begin{aligned}
&\sin \theta=f=\mu N=\mu m g \cos \theta \\
&\text { or } \tan \theta=\mu \Rightarrow \theta=\tan ^{-1}(\mu) \\
&\text { or } \tan \theta=\mu \Rightarrow \theta=\tan ^{-1}(\mu)
\end{aligned}
$$
(b) If angle of inclination is increased to $\alpha>\theta$. The angle of inclination of plane with horizontal it will slide down, then net force acting on the box, down the plane is
$$
\begin{aligned}
F_1 &=m g \sin \alpha-f=m g \sin \alpha-\mu N \\
&=m g \sin \alpha-\mu m g \cos \alpha \\
&=m g(\sin \alpha-\mu \cos \alpha) .
\end{aligned}
$$
(c) To keep the box either stationary or just move it up with uniform velocity $(a=0)$ upward force needced.
$$
\begin{aligned}
F_2 &=m g \sin \alpha+f \\
&=m g \sin \alpha+\mu N=m g \sin \alpha+\mu m g \cos \alpha \\
&=m g(\sin \alpha+\mu \cos \alpha)
\end{aligned}
$$
In this case, friction would act down the plane.
(d) The force applied $F_3$ to move the box with an upward acceleration $a$, then upward force needed.
$$
\begin{aligned}
&F_3-m g \sin \alpha-\mu m g \cos \alpha=m a \\
&F_3=m g(\sin \alpha+\mu \cos \alpha)+m a .
\end{aligned}
$$
For the box to just starts sliding down $(m g)$ the plane then
$$
\begin{aligned}
&\sin \theta=f=\mu N=\mu m g \cos \theta \\
&\text { or } \tan \theta=\mu \Rightarrow \theta=\tan ^{-1}(\mu) \\
&\text { or } \tan \theta=\mu \Rightarrow \theta=\tan ^{-1}(\mu)
\end{aligned}
$$
(b) If angle of inclination is increased to $\alpha>\theta$. The angle of inclination of plane with horizontal it will slide down, then net force acting on the box, down the plane is
$$
\begin{aligned}
F_1 &=m g \sin \alpha-f=m g \sin \alpha-\mu N \\
&=m g \sin \alpha-\mu m g \cos \alpha \\
&=m g(\sin \alpha-\mu \cos \alpha) .
\end{aligned}
$$
(c) To keep the box either stationary or just move it up with uniform velocity $(a=0)$ upward force needced.
$$
\begin{aligned}
F_2 &=m g \sin \alpha+f \\
&=m g \sin \alpha+\mu N=m g \sin \alpha+\mu m g \cos \alpha \\
&=m g(\sin \alpha+\mu \cos \alpha)
\end{aligned}
$$
In this case, friction would act down the plane.
(d) The force applied $F_3$ to move the box with an upward acceleration $a$, then upward force needed.
$$
\begin{aligned}
&F_3-m g \sin \alpha-\mu m g \cos \alpha=m a \\
&F_3=m g(\sin \alpha+\mu \cos \alpha)+m a .
\end{aligned}
$$
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.