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