Two blocks connected by a cord passing over a small, frictionless pulley rest on frictionless planes as shown. Find the acceleration of the blocks and the tension in the cord

Question

100 kg

30°

50 kg

53

Answer 1

First, we need to break down the forces acting on each block:

For the 100 kg block:
- Weight (W1) = mg = 100 kg * 9.8 m/s^2 = 980 N
- Tension in the cord (T) acting upwards
- Normal force (N1) acting perpendicular to the incline
- Friction force (f) opposing motion

For the 50 kg block:
- Weight (W2) = mg = 50 kg * 9.8 m/s^2 = 490 N
- Tension in the cord (T) acting downwards
- Normal force (N2) acting perpendicular to the incline
- Friction force (f) opposing motion

Next, we can break down the forces into components parallel and perpendicular to the incline for each block:

For the 100 kg block:
- Parallel force: 980 N * sin(30°) = 490 N
- Perpendicular force: 980 N * cos(30°) = 850 N

For the 50 kg block:
- Parallel force: 490 N * sin(30°) = 245 N
- Perpendicular force: 490 N * cos(30°) = 425 N

Now, we can calculate the acceleration of the blocks using Newton's second law:

Sum of forces = mass * acceleration

For the 100 kg block:
490 N - T = 100 kg * a
T = 490 N - 100 kg * a (equation 1)

For the 50 kg block:
T - 245 N = 50 kg * a
T = 50 kg * a + 245 N (equation 2)

Now, we can set the tension in the cord equal using equations 1 and 2:

490 N - 100 kg * a = 50 kg * a + 245 N
245 N = 150 kg * a
a = 1.63 m/s^2

Finally, we can calculate the tension in the cord using equation 1:
T = 490 N - 100 kg * 1.63 m/s^2
T = 324.7 N

Therefore, the acceleration of the blocks is 1.63 m/s^2 and the tension in the cord is 324.7 N.