Question

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.

Answer 1

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.