Question

A student attaches a rope to a 33 kg box, and drags it to the left with constant velocity of 1.11 m/s. The tension in the rope is 283 N at an angle of 33° to the ground.

How much does the box weigh?
323.4N

Find the x and y components of the applied (tension) force:

Fx =
237.34N

Fy =
154.13 N

How much friction must be present?

How much Normal force must be present?

Answer 1

Answer: To find the frictional force and normal force, we need to first find the gravitational force acting on the box, which is equal to its weight. We can find weight by multiplying the mass of the box by the acceleration due to gravity:

Weight = m * g

Weight = 33 kg * 9.8 m/s^2

Weight = 323.4 N

To find the x and y components of the tension force, we can use trigonometry:

Fx = Tension * cos(33°)

Fx = 283 N * cos(33°)

Fx = 237.34 N

Fy = Tension * sin(33°)

Fy = 283 N * sin(33°)

Fy = 154.13 N

Since the box is moving with constant velocity, we know that the net force on the box is zero. Therefore, the frictional force must be equal in magnitude and opposite in direction to the x component of the tension force:

Friction = -Fx

Friction = -237.34 N

The normal force is equal in magnitude and opposite in direction to the y component of the tension force:

Normal force = -Fy

Normal force = -154.13 N

Step-by-step explanation: