A box with a mass of 22kg is at rest on a ramp inclined at 45 degrees to the horizontal. The coefficient of friction between the box and the ramp are mu static = 0.78 mu kinetic = 0.65.
a) Determine the magnitude of the largest force that can be applied upward, parallel to the ramp, if the box is to remain at rest.
b) Determine the magnitude of the smallest force that can be applied onto the top of the box, perpendicular to the ramp, if the box is to remain at rest.
a. M*g = 22*9.8 = 215.6 N. = Wt. of box.
F = 215.6*sin45 = 152.5 N.
b. F = 215.6*cos45 =
To answer these questions, we need to analyze the forces acting on the box and use Newton's laws of motion. Let's break down the problem step-by-step.
a) To determine the largest force that can be applied upward, parallel to the ramp, without the box moving, we need to consider the forces acting on the box. In this case, the forces are the gravitational force (mg) acting downwards, the normal force (N) acting perpendicular to the ramp, and the static friction force (fs) acting parallel to the ramp.
We can start by calculating the normal force (N) and the gravitational force (mg). The normal force (N) is equal to the component of the gravitational force (mg) acting perpendicular to the ramp. In this case, since the ramp is inclined at 45 degrees, the normal force (N) can be calculated as:
N = mg * cos(45)
Next, we need to determine the maximum static friction force (fs) that can be exerted on the box to keep it at rest. The maximum static friction force (fs) is given by the equation:
fs = μs * N
where μs is the coefficient of static friction.
Now, we can substitute the values into the equation:
fs = 0.78 * N
Finally, we have determined the magnitude of the largest force that can be applied upward, parallel to the ramp, if the box is to remain at rest.
b) To determine the smallest force that can be applied onto the top of the box, perpendicular to the ramp, we need to consider the forces acting on the box. These forces are the gravitational force (mg) acting downwards, the normal force (N) acting perpendicular to the ramp, and the static friction force (fs) acting parallel to the ramp.
In this case, the smallest force that can be applied onto the top of the box, perpendicular to the ramp, would be when the box is on the verge of sliding downward. This occurs when the friction force (fs) reaches its maximum value, which is the kinetic friction force (fk).
The equation for the kinetic friction force (fk) is given by:
fk = μk * N
where μk is the coefficient of kinetic friction.
Now, we can substitute the coefficient of kinetic friction (μk) into the equation and calculate the magnitude of the smallest force that can be applied onto the top of the box, perpendicular to the ramp, if the box is to remain at rest.
It is important to note that in both cases, we assume the forces are applied parallel or perpendicular to the ramp.