Question

The two blocks m = 5 Kg and M = 25 Kg as shown in the figure are free to moves. The co-efficient of friction between the blocks is ${\mu }_s=0.4$ , but the co-efficient of friction between the ground and M is frictionless. What is the minimum horizontal force F required to hold m against M.?

Answer 1

Given that,

Mass $m=5kg$

Mass $M=25kg$

Coefficient $\mu =0.4$

Let F be the force which is required to hold m on M

$F=(m+M)a$

$a=\frac{F}{(m+M)}$

Now, the frictional force on m will be opposed by the pseudo force felt by it

Pseudo force on $m=m\times$acceleration of $(m+M)$

$ma=\mu N$

$ma=\mu mg$

$m\left(\frac{F}{m+M}\right)=\mu mg$

$F=\mu g(m+M)$

$F=0.4\times 10(5+25)$

$F=120N$

Hence, the minimum force is $120N$