Sol:

Given:

Mass, $$M_{1}$$ = 100kg

Mass, $$M_{2}$$ = 200kg

g = $$10m/s^{2}$$

The coefficient of friction between a and b,$$\mu _{1} = 0.2$$

The coefficient of friction between b and the ground, $$\mu _{2} = 0.3.$$

Frictional force acting between blocks A and B = $$f_{1} = \mu _{1}M_{1}g$$ $$\Rightarrow$$ 0.2 * 100 * 10

$$\Rightarrow$$200 N

Frictional force acting between blocks A and B = $$f_{2} = \mu _{2}(M_{1} + M_{2})g$$ $$\Rightarrow$$0.3 * (100 + 200) * 10

$$\Rightarrow$$0.3 * (300) * 10

$$\Rightarrow$$ 900 N

The minimum force required to move block B = $$f = f_{1} + f_{2}$$ $$\Rightarrow$$200 N + 900 N = 1100 N

Therefore, the minimum force required to move block B is 1100 N.

Explore more such questions and answers at BYJU’S.

Was this answer helpful?

0 (0)

(0)
(0)

#### Choose An Option That Best Describes Your Problem

Thank you. Your Feedback will Help us Serve you better.