Large boxes of pumpkins are set on a ramp to be unloaded out of a truck. The 300.0 lb boxes are set on the ramp which has an incline of 35 degrees. They must then be pushed to get started down the plane towards the bottom. If the coefficient of the static friction between the boxes and the ramp is .8, how much push applied parallel to the ramp is required to get one of the boxes of pumpkins moving?
To calculate the force required to get the box moving, we need to consider the forces acting on the box. In this case, the relevant forces are:
1. Weight (W): The force exerted by gravity on the box, which is equal to the mass of the box (300.0 lb) multiplied by the acceleration due to gravity (32.2 ft/s^2).
W = mass * gravity acceleration = 300.0 lb * 32.2 ft/s^2
2. Normal force (N): The perpendicular force exerted by the ramp on the box. It is equal to the weight of the box multiplied by the cosine of the incline angle (35 degrees).
N = W * cos(35 degrees)
3. Frictional force (Ff): The force of static friction between the box and the ramp. It is equal to the normal force (N) multiplied by the coefficient of static friction (.8).
Ff = N * coefficient of static friction
The force required to get the box moving is given by the equation:
F_push + Ff = W * sin(incline angle)
Rearranging the equation:
F_push = (W * sin(incline angle)) - Ff
Now, we can substitute the values into the equation:
F_push = (300.0 lb * 32.2 ft/s^2 * sin(35 degrees)) - (N * coefficient of static friction)
To calculate N, we use the equation:
N = W * cos(35 degrees)
Substituting the value:
N = (300.0 lb * 32.2 ft/s^2) * cos(35 degrees)
Substituting N and the coefficient of static friction back into the equation for F_push:
F_push = (300.0 lb * 32.2 ft/s^2 * sin(35 degrees)) - ((300.0 lb * 32.2 ft/s^2 * cos(35 degrees)) * 0.8)
Evaluating the equation will give you the force required to get the box moving.