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 7.5N parallel to the handle, what is the force of friction acting on the suitcase?
The force of friction equals the component of the pulling force along the floor, since there is no acceleration.
F(friction) = 7.5 N * cos45
To find the force of friction acting on the suitcase, we can use the concept of vector components.
First, let's break down the force you applied into its horizontal and vertical components. Since the angle between the handle and the horizontal direction is 45 degrees, the force you applied can be divided into two components:
1. The horizontal component: F_h = F * cos(theta)
where F is the force you applied (7.5 N) and theta is the angle between the handle and the horizontal direction (45 degrees).
F_h = 7.5 N * cos(45 degrees) = 7.5 N * 0.7071 ≈ 5.30 N
2. The vertical component: F_v = F * sin(theta)
F_v = 7.5 N * sin(45 degrees) = 7.5 N * 0.7071 ≈ 5.30 N
Since you are pulling the suitcase with a constant speed, the force of friction will be equal in magnitude but in the opposite direction to the horizontal component of the force you applied. Therefore, the force of friction acting on the suitcase is 5.30 N and it acts in the opposite direction of your applied force.