Question

(The figure model is attached below in a picture) For the figure above there is no friction between block A and the ramp. The mass of block A is 5.6kg. The ramp is 37.383198° from horizontal. Block A is being pulled up the ramp with a constant velocity by a line that is parallel to the ramp. What is the mass of block B?

Give a variable legend for this problem. The model for this problem:
mB (Mass of block B) =__________________________________ Answer________________________________

Answer 1

Forces on Block A:
Tension, normal reaction and sliding by gravity.
Forces on Block B:
Tension, and weight.

Here it is assumed that the string is light (negligible mass) and inextensible. Hence tension is equal on both blocks. Assumption 2: Block B moves down, so down is considered positive.

Sum of forces: (Block A)
T - mg sin theta = ma. ——— 1

Sum of forces: (Block B)
mg - T = ma. ——— 2
Acceleration is equal for both equations.

Adding 1 and 2,
mg - T + T - mg sin theta = (Total mass)a.

Let the mass of block B be x.

——> xg - 5.6g sin 37.383198 = (5.6 + x)a
——> 10x - 56 sin 37.383198 = (5.6 + x)a (Taking g as 10 m/s^2)
——> 10x - 56*3/5 = (5.6+x)a (Sin 37 can be taken as 3/5 from a triangle of sides 3,4,5)
——> 10x - 33.6 = (5.6+x)a
——> 10x - 33.6 = 0. (Acceleration is 0 as velocity is constant)
x = 3.6kg.