Question

Find the force required to move a load of 30 Newton up a rough inclined plane ,applied parallel to the plane. the inclination of plane is such that when the same body is kept on perfectly smooth plane inclined at an angle ,a force of 6 Newton applied at an inclination of 30 degree to plane keeps same equilibrium coefficient of friction between rough plane and load is 0.3.

Answer 1

Explanation:

sliding down is mgsinθ which is equal to 25 N here.

now, the component of the horizontal force F

a

along the incline plane should be equal to 25 N. For that,

F

a

cos30°=25

F

a

=

3

50

N

I HOPE IT HELP YOU

Answer 2

Answer:

The force required to move a load up a rough inclined plane is 13.96N.

Explanation:

Given the force on load up a rough inclined plane, W=mg = 30 N  

Force on body on perfectly smooth plane inclined, F = 6 N

Inclination of plane = 30degree

Coefficient of friction between rough plane and load $\mu= 0.3$        

The free body diagram for the forces acting on the body placed on inclined plane is given in the diagram.

• The weight, mg acting downwards is resolved into two components.
• One component mgcosθ normal to inclined surface and the other component mgsinθ is parallel to inclined surface
• And similarly the force at an angle 30º is resolved into F cos30,  parallel to inclined surface  and F sin30 normal to inclined surface.
• Normal reaction R is the contact force between object and inclined plane surface .      

At equilibrium, the normal force is $R = mg cos\theta - F sin30$

The frictional force, $\mu R = \mu ( mg cos\theta - F sin30 )$

The required force to keep the body at equilibrium on smooth surface   $= mg sin\theta$

thus we have,  $mg sin\theta = 6 cos30$

$\implies sin\theta = \frac{6 cos30}{mg}$    

$\implies sin\theta = \frac{6 \times \sqrt{\frac{3}{2} } }{30}$

$= \frac{\sqrt{3} }{10}= 0.17$              

And hence, $\theta = sin^{-1} (0.17) \approx 10^{o}$                                      

To move the load of 30N upward on rough inclined surface, minimum force required in parallel to inclined surface is

$F = mgsin\theta + \mu mg cos\theta$  

$= mg(sin\theta + \mu cos\theta )$  

$= 30 \times (sin10 + 0.3 cos10)$

$= 30 \times (0.17 + 0.3 \times 0.98)=50 N$

$= 30 \times 0.46=13.96 N$

Therefore, the force required to move a load up a rough inclined plane is 13.96N.