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If blocks A and B of mass 10 kg and 6 kg respectively, are placed on the inclined plane and released, determine the force developed in the link. The coefficients of kinetic friction between the blocks and the inclined plane are mA = 0.1 and mB = 0.3. Neglect the mass of the link.

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2 votes

Answer:

The force developed in the link is 6.36 N.

Step-by-step explanation:

Given that,

Mass of block A = 10 kg

Mass of block B = 6 kg

Coefficients of kinetic friction
\mu_(A)= 0.1

Coefficients of kinetic friction
\mu_(B)= 0.3

Suppose the angle is 30°

We need to calculate the acceleration

Using formula of acceleration


a=(m_(A)g\sin\theta+m_(B)g\sin\theta-\mu_(A)m_(A)g\cos\theta-\mu_(A)m_(A)g\cos\theta)/(m_(A)+m_(B))

Put the value into the formula


a=(10*9.8\sin30+6*9.8\sin30-0.1*10*9.8*\cos30-0.3*6*9.8*\cos30)/(16)


a=3.415\ m/s^2

We need to calculate the force developed in the link

For block A,

Using balance equation


ma=m_(A)g\sin\theta-\mu m_(A)g\cos\theta-T


T=ma+\mu m_(A)g\cos\theta-m_(A)g\sin\theta

Put the value into the formula


T=10*3.415+0.1*10*9.8*\cos30-10*9.8*\sin30


T=-6.36\ N

Negative sign shows the opposite direction of the force.

Hence, The force developed in the link is 6.36 N.

User Mattek
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