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posted by  eljose on 6/30/2008 8:54:14 PM  |  status: Live  

force and motion

Course Textbook Chapter Problem
Calculus Based Physics Fundamentals of Physics Extended (8th) by Resnick, Halliday, Walker 6 N/A
Question Details:
A loaded penguin sled weighing 70 N rests on a plane inclined at 20° to the horizontal (Figure 6-29). Between the sled and the plane the coefficient of static friction is 0.26, and the coefficient of kinetic friction is 0.16.


Figure 6-29

(a) What is the minimum magnitude of the force F, parallel to the plane, that will prevent the sled from slipping down the plane?
N
(b) What is the minimum magnitude F that will start the sled moving up the plane?
N
(c) What value of F is required to move the block up the plane at constant velocity?
N

 
help please

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(University of Texas at Tyler)
posted by Jordan Rey on 6/30/2008 9:06:23 PM  |  status: Live
Asker's Rating: Helpful   
eljose's comment:
"mmm i still don't get how u get (a)"
Response Details:
Normal force = m*g*sinθ = 24N
Frictional force = Normal*cosθ*coefficient of friction
(c) f = 24cos(20)*.16 = 3.6 N to keep it going at constant velocity
 
(b) f = 24cos(20)*.26 = 5.86 Newtons is the minimum force to start it moving up the plane.
 
 (a) 5.85 N  will keep it from moving if static.
 
Jordan Rey
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posted by suri.. on 6/30/2008 9:24:46 PM  |  status: Live
Asker's Rating: Lifesaver   
eljose's comment:
"thanks for the explanation"
Response Details:
(a) as we want the minimum force so that the sled doesnot slip down... the friction direction would be upwards and

f= mgsin20- .26mgcos20=6.839N

(b) as the sled moves up now the friction acts downwards

f = mgsin20 + .26mgcos20 = 41.04 N

(C) as it should move with a constant velocity the friction should be kinetic friction

f= mgsin20+.16mgcos20 = 34.46


here once the sliding starts we need 34.46N but to get to point of sliding we need 41.04 N
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