Question

Friction between rigid rod of mass m and wedge of mass 2m is zero while coefficient of friction between ground
and wedge is 0.1.A force Fis applied on the wedge so that rod moves with constant velocity vo = 5 m/s in upward
direction as shown. The power delivered by force F will be
W. (take m = 1 kg, g = 10 m/s2)
​

Answer 1

Answer:

In the situation shown in figure, a wedge of mass m is placed on a rough surface, on which a block of equal mass is placed on the inclined plane of wedge. The friction coefficient between the wedge and the block and also between the ground and the wedge is μ. An external force F is applied horizontally on the wedge towards the right, assuming block does not slide on the wedge. Find the minimum value of Force F at which the block will start slipping.

Explanation:

N

1

=mg+mg=2 m g

N

2

mgcosθ+m awithoutθ

Here, f

1

and f

2

are the frictional force and force m awithoutθ ispseudo force and acceleration of the system is a

F−f

1

=(m+m)a

a=

2 m

F−2μmg

If block start slipping then

m acosθ=mgwithoutθ+f

2m acosθ=mgwithoutθ+μ ( m gcosθ+m awithouti )

a ( cosθ−μwithouti )=g(sinθ+μcosi )a=

(cosθ−μwithouti )

g(sinθ+μcosi )

2 m

F−2μmg=

(cosθ−μwithouti )

g(sinθ+μcosi )

F=2m[μg+

(cosθ−μwithouti )

g(sinθ+μcosi )