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.
My work:
a. M*g = 22*9.8 = 215.6 N. = Wt. of box.
F = 215.6*sin45 = 152.5 N.
b. F = 215.6*cos45 = 152.45
is my work correct?
..... but where is your friction force?
It is to remain at rest both times, so use static coef = mu = 0.78
max friction force = mu m g cos 45 = mu *152.5 = 119
(because cos 45 = sin 45 = 0.707)
1. Push up
F = 152.5 + 119
2. push down adding to weight
152.5 = mu (m g).707 + .78 F = 119 + (.78) F
I agree.
a) Your work is correct. The weight of the box is equal to the force of gravity, which is given by the mass (22 kg) multiplied by the acceleration due to gravity (9.8 m/s^2). The weight of the box is 215.6 N. To determine the maximum force that can be applied upward parallel to the ramp, you use the following formula: F = m*g*sin(theta), where m is the mass of the box, g is the acceleration due to gravity, and theta is the angle of the ramp. In this case, the angle is 45 degrees. Plugging in the values, you get F = 215.6*sin(45) = 152.5 N.
b) Your work is also correct. To determine the minimum force that can be applied perpendicular to the ramp, you use the formula F = m*g*cos(theta). Again, plugging in the given values, you get F = 215.6*cos(45) = 152.45 N.
Your work is partially correct, but there are a few mistakes in your calculations.
a) To determine the magnitude of the largest force that can be applied upward, parallel to the ramp, if the box is to remain at rest, you need to consider the forces acting on the box. These forces include the weight of the box (mg), the normal force (N), and the frictional force (Ff).
The weight of the box is calculated correctly as W = mg = 22 kg * 9.8 m/s^2 = 215.6 N.
The normal force (N) is the perpendicular force exerted by the ramp on the box. In this case, the normal force is equal to the component of the weight perpendicular to the ramp. N = W * cos(45°) = 215.6 N * cos(45°) = 215.6 N * 0.7071 ≈ 152.7 N.
The frictional force (Ff) is given by the equation Ff = μ * N, where μ is the coefficient of friction and N is the normal force. In this case, the maximum static friction coefficient is given, so Ff = μ_static * N = 0.78 * 152.7 N ≈ 119.1 N.
To keep the box at rest, the maximum force that can be applied upward, parallel to the ramp, is equal to the maximum static friction force:
F_max = Ff_max = 119.1 N.
Therefore, the magnitude of the largest force that can be applied upward, parallel to the ramp, if the box is to remain at rest, is approximately 119.1 N.
b) To 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, you need to find the opposing force to the weight of the box. This force can be calculated as:
F_opposing = mg = 22 kg * 9.8 m/s^2 = 215.6 N.
Therefore, 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, is approximately 215.6 N.
So, in summary:
a) The magnitude of the largest force that can be applied upward, parallel to the ramp, if the box is to remain at rest, is approximately 119.1 N.
b) 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, is approximately 215.6 N.