UPS man pushes boxes up a ramp problem

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A UPS worker is tasked with pushing 120 kg boxes up a 32-degree ramp, needing to calculate the force exerted parallel to the floor to achieve an acceleration of 0.8 m/s². The coefficients of friction are 0.62 and 0.43. The challenge lies in adapting the existing formula for force parallel to the ramp to one that accounts for a horizontal push. The approach involves breaking down the forces into components parallel and perpendicular to the ramp, similar to how gravity is analyzed. Understanding this breakdown will facilitate the correct calculation of the required push force.
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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.
 
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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.)
 
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