IIT JEE , AIEEE Forum - Physics , Mechanics

Betu

Q.A block of mass m is placed on a trolley free to move on smooth horizontal surface. What maximum force can be applied on trolley so that block doesn't slide? Mass of troll is 2 m and co-efficient of friction between block and trolley is = 0.5

A	3 mg	B	3/2 mg
B	mg	A	mg/2

(This is same problem as being shown on the home page of GOIIT.com , the figure accompanying the problem unable to be shown..It can be taken from home page of GOIIT.com

hsbhatt

Consider the troll-block system. The mass is 3m and if the force applied is F, then a = F/3m.

In the reference frame of the troll, the forces acting on the block are the frictional force of 0.5mg (in the dirn of movement of the trolley) and the 'pseudo-force' of -ma. Since the block doesnt slide, ma = 0.5mg or F/3 = 0.5mg or F = 3mg/2 [Option B]

GaryK

Mass of trolley+block=3m
Force applied = F
Acceleration = F/(3m)

Considering only the block
Force acting backward= Pseudo force ma= m x F/(3m)
Force acting forward = frictional force (Friction acts forward since in a frictionless surface, the block would move backward) =

x m x g

In limiting case
Frictional force = Pseudo force
-->

x m x g = m x F/(3m)
-->

x m x g = F/3
--> F = 3 x

x m x g
--> F = 3 x 0.5 x 10m --> F = 3/2 mg

savvej

sorry to say but u all are worng.
i have already answered this question on the post:
http://www.goiit.com/posts/list/mechanics-question-on-force-37619.htm#186231

savvej

ans. is mg/2

iitkgp_bipin

Well done Savej.

Refer the link given by Savej. He has done it correctly.

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