Two people push on a crate. Person 1 pushes in the positive X direction with a force of 25N. The second person pushes in the negative X direction at an angle of 15 degrees below the horizontal with a magnitude of 18.5N the box moves with uniform velocity. What is the force of friction between the box and floor?
break up the negative x direction into components:
Vertical: 18.5sin15 upward
horizontal: 18.5cos15 in neg x direction
25-18.6cos15-friction=mass*acceleration=0
solve for friction force.
To find the force of friction between the box and the floor, we need to consider the net force acting on the box. Since the box moves with uniform velocity, it means that the net force on the box is zero.
Let's break down the forces acting on the box:
1. Force exerted by person 1 in the positive X direction: 25N
2. Force exerted by person 2 in the negative X direction: -18.5N (negative magnitude indicates opposite direction)
- We also need to determine the horizontal component of this force since it is applied at an angle below the horizontal. The horizontal component can be calculated by multiplying the magnitude of the force by the cosine of the angle:
Horizontal component of force 2 = 18.5N * cos(15°)
Now, let's sum up the forces:
Net force in the X direction = Force 1 + Horizontal component of Force 2
Net force in the X direction = 25N + 18.5N * cos(15°)
Since the box moves with uniform velocity, the net force in the X direction is balanced by the force of friction acting in the opposite direction:
Force of friction = -(25N + 18.5N * cos(15°))
Evaluating this expression will give you the force of friction between the box and the floor.