Calculate Frictional Forces Needed for 6.5kg Shopping Cart on 13° Incline

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SUMMARY

The discussion focuses on calculating the horizontal force required to push a 6.5kg shopping cart up a 13° incline with an acceleration of 1.61 m/s². The user applies Newton's second law and resolves forces, but encounters difficulty due to the presence of two variables: the coefficient of friction and the normal force. The correct approach involves calculating the gravitational force component along the incline and incorporating the frictional force, which requires knowing the coefficient of friction.

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  • Understanding of Newton's second law of motion
  • Basic knowledge of trigonometry for resolving forces
  • Familiarity with friction concepts, including the coefficient of friction
  • Ability to calculate normal force on an incline
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wind522
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I have a question..

A shopper pushes a 6.5kg shopping cart up a 13 degree incline, heading east. Find the magnitude of the horizontal force, F, needed to give the cart an acceleration of 1.61 m/s2 (seconds squared).

This is what I have done.
sigmaFx= F(what I'm trying to find) - Fg,x - Ff
MAnet = F - mgsin(theta) - u(mieu)Fn
F = MAnet + mgsin(theta) + uFn
F = (6.5kg)(1.61m/s2) + (6.5kg)(9.81m/s2)sin13 +u(63.13070718N)
And here's where I'm stuck because there's two variables. Does anyone know what I'm doing wrong?

The answer should be in Newtons
 
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Are you supposed to be including friction? If so, you'll need the coefficient of friction.

Also, note that the applied force is horizontal, thus if you are finding components parallel to the incline you must take that into account.
 

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