Question
The two blocks, $$m = 10\,kg$$ and $$M = 50\,kg$$ are free to move as shown. The coefficient of static friction between the blocks is 0.5 and there is no friction between $$M$$ and the ground. A minimum horizontal force $$F$$ is applied to hold $$m$$ against $$M$$ that is equal to
The two blocks, $$m = 10\,kg$$ and $$M = 50\,kg$$ are free to move as shown. The coefficient of static friction between the blocks is 0.5 and there is no friction between $$M$$ and the ground. A minimum horizontal force $$F$$ is applied to hold $$m$$ against $$M$$ that is equal to
A.
$$100\,N$$
B.
$$50\,N$$
C.
$$240\,N$$
D.
$$180\,N$$
Answer :
$$240\,N$$
Solution :
As $$m$$ would slip in vertically downward direction, then
$$\eqalign{ & mg = \mu N \cr & \Rightarrow N = \frac{{mg}}{\mu } = \frac{{10.0}}{{0.5}} = 200\,Newton \cr} $$
Same normal force would accelerated $$M,$$
thus $${a_M} = \frac{{200}}{{50}} = 4\,m/{s^2}$$
Taking $$m+ M$$ as system
$$F = \left( {m + M} \right)4 = 240\,N$$
As $$m$$ would slip in vertically downward direction, then
$$\eqalign{ & mg = \mu N \cr & \Rightarrow N = \frac{{mg}}{\mu } = \frac{{10.0}}{{0.5}} = 200\,Newton \cr} $$
Same normal force would accelerated $$M,$$
thus $${a_M} = \frac{{200}}{{50}} = 4\,m/{s^2}$$
Taking $$m+ M$$ as system
$$F = \left( {m + M} \right)4 = 240\,N$$
