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.

The force developed in the link is 6.36 N.

Explanation:

Given that,

Mass of block A = 10 kg

Mass of block B = 6 kg

Coefficients of kinetic friction

Coefficients of kinetic friction

Suppose the angle is 30°

We need to calculate the acceleration

Using formula of acceleration

Put the value into the formula

We need to calculate the force developed in the link

For block A,

Using balance equation

Put the value into the formula

Negative sign shows the opposite direction of the force.

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

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west