
#1
Apr1504, 05:58 PM

P: 44

This question is driving me insane please help.
A 25 kg block of ice slides down an incline 2.5 m long and 1.5 m high. A man pushes up the ice parallel to the incline so it slides down at a constant speed. The coefficient of friction between the ice and incline is 0.10. Find the force exerted by the man. 



#2
Apr1504, 06:15 PM

Mentor
P: 40,880

Start by identifying all the forces on the block of ice. I count four. 



#3
May1205, 02:13 AM

P: 30

ok I have the same problem.
The four forces I have found are: Fg = 147.1N (directly down) FN = 195.9N (perpendicular to the slope) Ff = 19.6N (up parallel the slope) F = 147.1N (down parallel the slope) just want to make sure here.. now I add the 4 vectors together and find out what Force I need to add (up parallel to the slope) in order to make the net force = 0 ? thanks 



#4
May1205, 02:53 AM

P: 58

Force and Friction on an Inclined PlaneThere are trig functions that should be applied here because the block is at an incline of [tex]\theta[/tex]. Once you draw it, you should have something like:: [tex] \Sigma F_{x} = F_{g}\sin\theta + F_{push} + f_{k} = ma[/tex] [tex] \Sigma F_{y} = n  F_{g}\cos\theta = 0[/tex] 



#5
May1205, 08:23 AM

Mentor
P: 40,880

(1) weight = mg, acting down. This will have components perpendicular and parallel to the incline. If you call the angle that the incline makes with the horizontal [itex]\theta[/itex], then [itex]W_{parallel} = mg \sin \theta[/itex] down the incline and [itex]W_{perpendicular} = mg \cos \theta[/itex] into the incline.To find the applied force, do as PhysicsinCalifornia advised: set the net force parallel to the plane equal to zero: [itex]mg \sin \theta  \mu mg \cos \theta  F = 0[/itex]. 


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