Solving a Friction Problem with a 32° Ramp

[egrus8]

Can someone help with this problem I don't even know where to start I have my freebody diagram drew up but I don't know where to go from there

A UPS worker is loading a truck by pushing boxes up a ramp that makes an angle of 32 degrees above the floor. The coefficients of friction between a box and the ramp are 0.62 and 0.43.

A) What is the minimum push up the ramp that the worker must exert on the 100 box to keep the box at rest? Hint: First draw a freebody diagram for the box without the worker's push and then add the push.

B) What push must the worker exert parallel to the floor (not the plane) to cause the box to move up the plane at a constant speed 0. 2

 

[Rake-MC]

I am going to assume that 0.62 is static friction and 0.43 is kinetic friction.

The '100 box' is 100kg?
The forces acting on the box are gravity, normal reaction, friction and the worker pushing.

Force due to gravity = mg.

The component parallel to the inclined plane that is going to make it harder for the worker is therefore going to be mgsin(32).

The normal reaction force is going to be perpendicular to the inclined plane which will be equal to mgcos(32).

Multiply this by the co-efficient of static friction to get .62mgcos(32).

To remain stationary he must push enough force to get .62mgcos(32) and mgsin(32) to be equal.

 

[egrus8]

yes you assumed right. but I am still unsure as to how to solve this

 

[cstcc09]

Any follow up on this? I have the same problem but 70kg instead of 100kg. I set up my equation:

Fcos(32) = .62[Fsin(32)+wcos(32)]+wsin(32)
w = mg = weight which for me would be 70(9.8) = 686N

So I get... .848F = .329F + 360.7 + 363.5

Then I solved for F and got F=1395N, which is definitely NOT the correct answer...

Any help would be much appreciated.

 

[asteriatic]

sin 32 degrees

I would first gather all the necessary information and data to solve this problem. This includes the angle of the ramp, the coefficients of friction, and the weight of the box (assuming it is 100 kg).

To solve part A, we can use the concept of equilibrium, where the forces acting on the box are balanced. The forces acting on the box are its weight (mg) and the normal force (N) from the ramp. We can also add the force of friction (Ff) in the opposite direction of motion.

Using the coefficient of friction (μ), we can calculate the maximum force of friction (Ffmax) as μN. We can then use trigonometry to find the normal force (N) as N = mgcosθ, where θ is the angle of the ramp.

To keep the box at rest, the worker must exert a force (Fp) equal to the force of friction in the opposite direction, so Fp = Ffmax = μN. Substituting our values, we get Fp = μ(mgcosθ). Solving for Fp, we get a minimum push of approximately 564.8 N.

For part B, we need to consider that the box is now moving at a constant speed, so the forces are balanced once again. This time, the forces acting on the box are its weight (mg), the normal force (N) from the ramp, and the force of the worker's push (Fp) parallel to the floor.

Using the angle of the ramp and the given speed, we can find the component of gravity acting parallel to the ramp as mgcosθ. This is balanced by the force of the worker's push (Fp) and the force of friction (Ff) in the opposite direction.

Using the coefficient of friction, we can find the force of friction as Ff = μN. Substituting our values, we get μ(mgcosθ) + Fp = μN. Solving for Fp, we get a required push of approximately 64.7 N.

In summary, to solve this friction problem, we used the concepts of equilibrium, trigonometry, and the coefficients of friction. It is important to draw a freebody diagram to visualize the forces acting on the box and to carefully apply the equations to solve the problem.