Question

05.
In figure, A is 10 kg block and B is 5 kg block. The
minimum value of mass of C be placed on A to
prevent it from sliding:
025
(a) 5 kg
10 kg (c) 20 kg (d) 40 kg​

Answer 1

Answer:

I think the answer is 5 kg

Answer 2

Answer: According to me it should be 10kg

Explanation:The force(tension) exerted by B on string = mxg =50N

Static friction on block A to keep it in place = fs =μ(M+x)g

x=mass of C

fs=tension in string = 50N

50=(0.25).(10+x).10

X=10 kg