a 5kg block, resting on a rough horizontal table is connected by a light in extensible string passing over a light frictionless pulley to another block of mass 2 kg. the 2 kg block hangs vertically. a force of 60N is applied to the 5 kg block at an angle of 10° to the horizontal, causing the block to accelerate to the left.

Question

a) calculate the magnitude of the horizontal component of the 60N force

Answer 1

To calculate the horizontal component of the 60N force, we first need to find the total force acting on the 5 kg block in the horizontal direction.

The total force acting on the 5 kg block in the horizontal direction can be calculated as the sum of the horizontal component of the applied force and the tension in the string:

Total horizontal force = Horizontal component of the applied force - Tension in the string

First, let's find the horizontal component of the applied force:

Horizontal component of the applied force = Force * cos(angle)
= 60N * cos(10°)
= 60N * 0.9848
≈ 59.09 N

Now, let's find the tension in the string. Since the 2 kg block is hanging vertically and accelerating upwards, the tension in the string is equal to the total force acting on the 2 kg block:

Tension = Mass * acceleration
= 2 kg * g
= 2 kg * 9.81 m/s^2
≈ 19.62 N

Now, we can calculate the total horizontal force acting on the 5 kg block:

Total horizontal force = 59.09 N - 19.62 N
= 39.47 N

Therefore, the magnitude of the horizontal component of the 60N force is approximately 39.47 N.