Block A with mass 4kg, that is at rest on a rough horizontal table, is connected to another block B with mass 8kg by a light inextensible string passing over a frictionless pulley.
A force F1 of magnitude 96 N is applied vertically upwards on block B as shown in the diagram below
A force F2 of magnitude 75 N is now applied at an angle of 25 degrees with the horizontal on block A and the block accelarates to the left. The kinectic friction force on block B is 11.76N. ignore the effects of air friction.
1. state Newto's second law of motion in words
2. draw a labelled free-body diagram for block A
3. calculate the magnitude of the:
a) accelaration of bloclk A
b) Tension in the rope connected to block A
1. Newton's second law of motion states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass.
2. Free body diagram for block A:
- Tension force (T) acting to the right
- Force due to F2 (75 N) at an angle of 25 degrees with horizontal, broken down into horizontal (F2cos25) and vertical (F2sin25) components
- Kinetic friction force (11.76 N) acting to the right
3.
a) To calculate the acceleration of block A, we need to consider the net force acting on it.
Net force = Tension force - Kinetic friction force - Component of force due to F2 in the opposite direction of motion
Net force = T - 11.76 - F2cos25
Using Newton's second law:
Net force = mass x acceleration
T - 11.76 - F2cos25 = (mass of block A) x acceleration
T - 11.76 - 75cos25 = 4a
T - 11.76 - 67.89 = 4a
T - 79.65 = 4a
T = 4a + 79.65
b) We also know that the tension force in the rope connected to block A is equal to the weight of block A plus the force due to F2sin25:
T = Weight of block A + F2sin25
T = (mass of block A) x (acceleration due to gravity) + 75sin25
T = 4 x 9.81 + 75sin25
T ≈ 39.24 + 31.46
T ≈ 70.7 N
Therefore:
a) The acceleration of block A is 3.415 m/s^2
b) The tension in the rope connected to block A is approximately 70.7 N