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 a=0.1 and b=0.3. neglect the mass of the link. ( the incline is at 30 degrees up from the -x axis, the link connects block a and b and block a is futher up the incline than block b)

The force developed in the link is 6.37N.and this can be determined by forming the equation of motion in the vertical and horizontal direction.

Given :

Blocks a and b of mass 10 kg and 6 kg respectively, are placed on the inclined plane and released.
The coefficients of kinetic friction between the blocks and the inclined plane are a=0.1 and b=0.3.
Neglect the mass of the link.

Equation of motion is formed in order to determine the force developed in the link.

Equation of motion for block A in the vertical direction.

[tex]\rm N_A - 10\times 9.81 \times cos30^\circ = 10\times 0[/tex]

[tex]\rm N_A = 10\times 9.81 \times cos30^\circ[/tex]

[tex]\rm N_A = 84.96 \; N[/tex]

Equation of motion for block A in the horizontal direction.

[tex]\rm 10\times 9.81 \times sin30^\circ -0.1\times 84.96 -F = 10a[/tex]

49.05 - 8.496 - F = 10a

40.554 - F = 10a

[tex]\rm \dfrac{40.554-F}{10}=a[/tex] ---- (1)

Equation of motion for block B in the vertical direction.

[tex]\rm N_B - 6\times 9.81 \times cos30^\circ = 6\times 0[/tex]

[tex]\rm N_B = 6\times 9.81 \times cos30^\circ[/tex]

[tex]\rm N_B = 50.97 \; N[/tex]

Equation of motion for block B in the horizontal direction.

[tex]\rm 6\times 9.81 \times sin30^\circ -0.3\times 50.97 +F = 6a[/tex]

F + 14.14 = 6a ---- (2)

Now, substitute the value of 'a' in equation (2).

[tex]\rm F + 14.14 = 6\times \dfrac{40.554-F}{10}[/tex]

5F + 70.7 = 121.662 - 3F

8F = 50.962

F = 6.37 N

So, the force developed in the link is 6.37N.

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