A block of mass $m$ is kept on a plank. The coefficient of friction between the plank and block is 1. The plank is slowly raised from one end so that it makes an angle with horizontal. the forces of friction acting on the plank, when $\theta =30{}^\circ $ and $\theta =60{}^\circ $ are respectively
A. $\dfrac{mg}{2},\dfrac{mg}{2}$
B. $\dfrac{\sqrt{3}}{2}mg,\dfrac{mg}{2}$
C. $\dfrac{mg}{2},\dfrac{\sqrt{3}}{2}mg$
D. $mg,mg$
Answer
635.7k+ views
Hint: Here, we need to find the expression of the frictional force. The expression can be found by resolving the forces acting on the body into their components. Finally, we will equate the frictional force with the balancing force to find the answer.
Formula used:
$\alpha ={{\tan }^{-1}}(\mu )$
$f=mg\sin \theta $
Complete answer:
Here, the different types of forces acting on the box can be visualized with the help of the following diagram:
Also, in the question it is given that the coefficient of friction between the body and the plank is 1. This can be written as:
$\mu =1$
Hence, the angle of repose can be found using equation: $\alpha ={{\tan }^{-1}}(\mu )$ ---(i)
Hence, here, $\alpha ={{\tan }^{-1}}(1)=45{}^\circ $
We know that as long as the angle of inclination of the plank remains below the angle of repose, and the body is at rest, the expression for the force of friction acting on the body is given by:
$f=mg\sin \theta $ ------(ii)
Hence, for $\theta =30{}^\circ $:
Frictional force acting on the body = ${{f}_{1}}=mg\sin 30$
$\Rightarrow {{f}_{1}}=mg\times \dfrac{1}{2}=\dfrac{mg}{2}$
Also, as soon as the angle of the plank increases beyond the angle of repose, the expression for the frictional force becomes:
$f=mg\cos \theta $ --------(iii)
Hence, for , $\theta =60{}^\circ $:
Frictional force acting on the body =${{f}_{2}}=mg\cos 60$
$\Rightarrow {{f}_{2}}=mg\times \dfrac{1}{2}=\dfrac{mg}{2}$
Hence, in both the cases, i.e., in case of $\theta =30{}^\circ $ and $\theta =60{}^\circ $, the frictional force acting on the body is the same.
So, the correct answer is “Option A”.
Note:
Write the expression for the frictional force carefully. The frictional force acting on the body kept on a plank depends on the inclination of the plank. The expressions for frictional force are different and depend upon whether the angle of inclination is less or more than the angle of repose.
Formula used:
$\alpha ={{\tan }^{-1}}(\mu )$
$f=mg\sin \theta $
Complete answer:
Here, the different types of forces acting on the box can be visualized with the help of the following diagram:
Also, in the question it is given that the coefficient of friction between the body and the plank is 1. This can be written as:
$\mu =1$
Hence, the angle of repose can be found using equation: $\alpha ={{\tan }^{-1}}(\mu )$ ---(i)
Hence, here, $\alpha ={{\tan }^{-1}}(1)=45{}^\circ $
We know that as long as the angle of inclination of the plank remains below the angle of repose, and the body is at rest, the expression for the force of friction acting on the body is given by:
$f=mg\sin \theta $ ------(ii)
Hence, for $\theta =30{}^\circ $:
Frictional force acting on the body = ${{f}_{1}}=mg\sin 30$
$\Rightarrow {{f}_{1}}=mg\times \dfrac{1}{2}=\dfrac{mg}{2}$
Also, as soon as the angle of the plank increases beyond the angle of repose, the expression for the frictional force becomes:
$f=mg\cos \theta $ --------(iii)
Hence, for , $\theta =60{}^\circ $:
Frictional force acting on the body =${{f}_{2}}=mg\cos 60$
$\Rightarrow {{f}_{2}}=mg\times \dfrac{1}{2}=\dfrac{mg}{2}$
Hence, in both the cases, i.e., in case of $\theta =30{}^\circ $ and $\theta =60{}^\circ $, the frictional force acting on the body is the same.
So, the correct answer is “Option A”.
Note:
Write the expression for the frictional force carefully. The frictional force acting on the body kept on a plank depends on the inclination of the plank. The expressions for frictional force are different and depend upon whether the angle of inclination is less or more than the angle of repose.
Recently Updated Pages
The given figure shows two endocrine glands marked class 11 biology NEET_UG

Match columnI with columnII and select the correct class 11 biology NEET

Match column I with column II and select the correct class 11 biology NEET_UG

Which floral family has left 9 right + 1 arrangement class 11 biology NEET_UG

Which is not a variety of sheep A Lohi B Beetal C Nellore class 11 biology NEET_UG

Match column I with column II and select the correct class 11 biology NEET_UG

Trending doubts
Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Two of the body parts which do not appear in MRI are class 11 biology CBSE

1 ton equals to A 100 kg B 1000 kg C 10 kg D 10000 class 11 physics CBSE

10 examples of law on inertia in our daily life

