The blocks in the system illustrated below are moving such that block A is moving up the ramp. the coefficient of kinetic friction between the ramp and the sliding block is 0.11. determine the magnitude of acceleration of the blocks below and the magnitude of force of tension in the string.
We can begin by drawing a free body diagram for each block:
For block A:
- Weight (mg) pointing downwards
- Normal force (N) perpendicular to the ramp
- Force of kinetic friction (fk) pointing opposite to the direction of motion
- Tension force (T) pointing upwards
For block B:
- Weight (Mg) pointing downwards
- Tension force (T) pointing upwards
We can use Newton's second law (F=ma) to write equations for each block in the vertical and horizontal directions:
For block A in the vertical direction:
T - mg = ma (1)
For block B in the vertical direction:
Mg - T = Ma (2)
For block A in the horizontal direction:
fk = ma (3)
To solve for the acceleration, we can eliminate T from equations (1) and (2) by substituting equation (1) into equation (2):
Mg - (ma + mg) = Ma
Simplifying:
a = g(M - m)/(M + m) = 2.96 m/s^2
To solve for the tension force, we can substitute the value of acceleration (a) from equation (3) into equation (1) to get:
T - mg = fk
T = fk + mg = (0.11)(9.81)(2.5) + (0.5)(2.5) = 3.60 N
Therefore, the acceleration of the blocks is 2.96 m/s^2 and the magnitude of tension force in the string is 3.60 N.