You are pulling a suitcase through the airport at a constant speed. The handle of the suitcase makes an angle of 45 degrees with respect to the horizontal direction. If you pull with a force of 7.5 N parallel to the handle, what is the force of friction acting on the suitcase?

Wouldn't the cosine function apply here?

To find the force of friction acting on the suitcase, we need to use the given angle and force information.

First, let's resolve the applied force into its vertical and horizontal components. Since the angle between the handle and the horizontal direction is 45 degrees, the horizontal component of the force is equal to the force magnitude multiplied by the cosine of the angle:

Horizontal component = 7.5 N * cos(45°) ≈ 7.5 N * 0.707 ≈ 5.30 N

The vertical component of the force can be found in the same way, but by multiplying the force magnitude by the sine of the angle:

Vertical component = 7.5 N * sin(45°) ≈ 7.5 N * 0.707 ≈ 5.30 N

Since the suitcase is moving at a constant speed, the net force in the horizontal direction must be zero. The force of friction is responsible for balancing the horizontal component of the applied force.

Thus, the force of friction acting on the suitcase is equal in magnitude but opposite in direction to the horizontal component of the applied force:

Force of friction = -5.30 N

Note that the negative sign indicates that the force of friction acts in the opposite direction to the applied force.

To find the force of friction acting on the suitcase, we need to consider the forces at play in this scenario. There are two main forces involved: the force you apply parallel to the handle and the force of friction opposing the motion of the suitcase.

We can break down the force you apply into two components: one parallel to the horizontal direction and one perpendicular to it. Since the handle makes an angle of 45 degrees with respect to the horizontal direction, the force you apply can be divided into two components: one in the horizontal direction and one in the vertical direction.

The component of the force parallel to the handle is given by the formula:

Force_parallel = Force * cos(angle)

Plugging in the values, we have:

Force_parallel = 7.5 N * cos(45 degrees)
= 7.5 N * 0.7071
≈ 5.30 N

Now, the force of friction is equal in magnitude but opposite in direction to the component of the force parallel to the handle:

Frictional force = -Force_parallel

Therefore, the force of friction acting on the suitcase is approximately 5.30 N.