Question

Positions of 2 blocks are placed on the plane and released as depicted in the figure. The coefficients of kinetic friction between the blocks and the ground are uA=0.2 and us=0.4, respectively. Find the force emerging in the link which connects blocks A and B together. Mass of the link is neglected. Block A and block B weighs 20kg and 4 kg.

Answer 1

Let the force in the link is T and inclination of plane with ground is $\theta$.

From Free Body Diagram of A, equation of motion is...

$$m_Ag\sin\theta-\mu_Amg\cos\theta-T=ma\\ 20g\sin\theta-0.2\times 20 g \cos\theta-T=20a....Eq[1]$$

From Free Body Diagram of B, equation of motion is...

$$4g\sin\theta-0.4\times4g\cos\theta+T=4a\\$$

now multiplying above equation by "5", we have...

$$20g\sin\theta-0.4\times 20 g \cos \theta + 5T=20a....Eq[2]$$

substracting Eq[1] by Eq[2], we have...

$$T=\cfrac{4g\cos\theta}{6}$$

Hence emerging force in the link is $\cfrac{4g\cos\theta}{6}.$