What is the required force to push a box up a slippery ramp?

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SUMMARY

The discussion focuses on calculating the force required to push a 10 kg box up a slippery ramp with a coefficient of friction of μ = 0.1. The initial calculations mistakenly identified the normal reaction and frictional forces, leading to an incorrect force of 10N instead of the correct 27N. Participants emphasized the importance of drawing a force diagram and applying Newton's second law to accurately determine the forces acting on the box.

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Homework Statement


A 10 kg
box is being pushed up a slippery ramp as shown in Figure 2.1.
The coefficient of friction between the box and the ramp is just
μ = 0.1.
(a) What force does the man need to apply to the box to
keep it traveling up the ramp at a steady speed?

Homework Equations


I have attached figure 2.1

The Attempt at a Solution


I worked out the normal reaction by finding the vertical component of the normal reaction acting on the box (as =mg) then i found out fmax by fmax = uN
but my answer is 10N and the actual answer is 27N!
what did i do wrong
 

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luysion said:
I worked out the normal reaction by finding the horizontal component of the normal reaction...

Are you sure you meant to say 'normal reaction' twice here? I'm a bit confused by your work; what are you taking the horizontal component of?

Also, what do you mean with fmax? Is that the force the person has to push?
 
for this problem you'll want to start out by drawing a force diagram for the box and then applying Newton's second law.
 
fmax = frictional force
can someone please show me how to do this its bugging the crap out of me,
sonic91 arent the forces acting on the box friction and the normal reaction

mintea;
the normal reaction is angles because of the inclination, so the upwards force on the box i.e. that acts directly vertically would be the vertical component of the normal reaction i bymistake wrote the horizontal opps heh cheers
 
Last edited:
In that case, your work looks right; you just may be missing a force the man has to work against... I suggest what sonic91 said: put everything on a force diagram and go from there.
 

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