Question

A $5.0kg$ crate is on an incline that makes an angle of ${30}^{°}$ with the horizontal. If the coefficient of static friction is $0.5$, the maximum force that can be applied parallel to the plane without moving the plane to hold the crate at rest is

• A. Zero
• B. $3.3N$
• C. $30N$
• D. $46N$

Answer 1

The correct option is B

$3.3N$

Step 1: Given data

Mass of the crate is $\mathrm{m}=5.0\mathrm{kg}$

The crate is inclined at an angle $\theta ={30}^{°}$

Coefficient of friction $\mu =0.5$

The diagrammatic representation of the given data is given below:

Step 2: Determine the force acting down the plane due to the weight

Since we know that there is a force acting on the body due to the weight in the downward direction whose horizontal component is $mg\sin \theta$ and the vertical component is $mg\cos \theta$.

Another normal force $N$ will be acting just normal to the plane and frictional force $f$ is also acting just opposite to the $mg\sin \theta$ up the plane.

Force acting down the plane due to the weight will be $mg\sin \theta$.

$\Rightarrow 5\times 9.8\times \sin {30}^{°}$

$\Rightarrow 24.5N$

Step 3: Determine the maximum force that can be applied parallel to the plane without moving the plane

The frictional force acting up the plane,

$f=\mu N=\mu mg\cos \theta$

On putting the required values in the above equation, we get,

$\Rightarrow f=0.5\times 9.8\times \cos {30}^{°}$

$\Rightarrow f=21.217N$

Thus, the maximum force acting on the plane to hold the crate at rest

$\Rightarrow 24.5-21.217\simeq 3.3N$

Hence, option B is the correct answer.