Find out the velocity of the system:
Blocks A and B of mass 20 kg and 45 kg, respectively are associated by a weightless rope over a frictionless pulley as shown in Figure a). Suppose a coefficient of friction of 0.2 for all the planes. Find out the velocity of the system 5 seconds after starting from rest.
Solution
Consider T be the tension in the strings. This is not rather easy to determine the direction of motion. Let us suppose that block B is moving down the plane such that block A climbs up the plane. It gives us the direction of acceleration for each of block and also the directions of frictional forces acting on each block.
Refer Figures (b) and (c). These figures give us free-body diagram for each block.
Consider Block A
We may write the following equilibrium equations. Supposing x axis along the plane and y axis perpendicular to the plane.
∑ F_{x } = 0, ∴ T = 20 a + 0.2 N_{1 } + 20 g sin 30^{o} ---------- (1)
∑ F_{y} = 0, N_{1} = 20 g cos 30^{o} = 169.91 N ------------ (2)
By Substituting in (1), we obtain
T = 20 a + 132.08 -------- (3)
Consider Block B
Now we may writer equilibrium equations for this block.
∑ F_{x} = 0, T + 0.2 N _{2} + 45 a = 45 g sin 60^{o} --------- (4)
∑ F_{y} = 0, N _{2} = 45 g cos 60^{o}
= 220.72 N ----------- (5)
By Substituting in (4), we obtain
T + 44.145 + 45 a = 382.30
∴ T + 45 a = 338.155 --------- (6)
Solving Eq. (3) and Eq. (6), we obtain
T = 20 a + 132.08 = - 45 a + 338.155
∴ 65 a = 206.07
∴ a = 3.17 m / sec^{2}.
The value of a, we have attained is positive and our supposition that block B moves down the plane is correct.
We can now apply kinematic equation,
v = v_{o} + a t ∴ v = 0 + 3.17 × 5
= 15.85 m / sec ^{2}.