UPS man pushes boxes up a ramp problem

[egrus8]

Could someone help with this problem I don't know how to set it up with the worker pushing parallel to the floor instead of the plane.

A UPS worker is loading a truck by pushing boxes weighing 120 kg 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.

What push must the worker exert parallel to the floor (not the plane) to cause the box to accelerate up the plane at an acceleration of 0.8

the formula for parallel to the plane is P=ma + mgsin(θ)+μ(mgcos(θ) for the push, the part I am having trouble with is how to get the formula for parallel to the floor instead of the plane? Any help will be truly appreciated.

 

[alphysicist]

Hi egrus8,

You can still write the equation for the components parallel to the plane, and perpendicular to the plane. The problem is just saying that the applied force (from the worker) is horizontal.

You can treat it in a similar way that you treat the vertical force of gravity. (You can think of the gravity force as being perpendicular to the floor, and you found its components that are parallel and perpendicular to the plane for the force equations.)

 

[baba89]

Thank you for reaching out for assistance with this problem. I am happy to help you understand and set up the problem correctly.

First, it is important to clarify the difference between "parallel to the floor" and "parallel to the plane" in this context. When we say "parallel to the floor," we are referring to the direction of the force that the UPS worker is exerting on the box. In this case, the worker is pushing the box horizontally, parallel to the floor. On the other hand, "parallel to the plane" refers to the direction of the ramp itself, which is at an angle of 32 degrees above the floor.

To set up the problem correctly, we need to use the formula for the net force in the direction of motion, which is given by F = ma. In this case, the acceleration is given as 0.8 m/s^2. We also need to consider the forces acting on the box, which are the push from the worker, the force of gravity (mg), and the frictional force from the ramp.

To find the push force parallel to the floor, we can use trigonometry to break down the forces into their components. The push force can be broken down into two components: one parallel to the plane (Fpar) and one perpendicular to the plane (Fperp). The force of gravity can also be broken down into two components: one parallel to the plane (mgcosθ) and one perpendicular to the plane (mgsinθ).

Using this information, we can set up the equation for the net force in the direction of motion (F = ma) as follows:

Fpar - mgcosθ - μ(mgsinθ) = ma

We can rearrange this equation to solve for the push force parallel to the floor (Fpar):

Fpar = ma + mgcosθ + μ(mgsinθ)

Plugging in the given values for mass (m = 120 kg), acceleration (a = 0.8 m/s^2), angle (θ = 32 degrees), and coefficients of friction (μ = 0.62 and 0.43), we can solve for Fpar:

Fpar = (120 kg)(0.8 m/s^2) + (120 kg)(9.8 m/s^2)(cos 32) + (0.62)(120 kg)(9.8 m/s^2)(sin