Figure One: The Penguin Problem A loaded penguin sled weighing 80 N rests on a p
ID: 1583095 • Letter: F
Question
Figure One: The Penguin Problem A loaded penguin sled weighing 80 N rests on a plane inclined at an angle 20° to the horizontal Figure One). Between the sled and the plane, the coefficient of static friction is 0.25, and the coefficient of kinetic friction is 0.15. (a) What is the least magnitude of the force F, parallel to the plane which will prevent the sled from slipping down the plane? (b) W will start the sled moving up the plane? (c) W constant velocity? hat is the minimum magnitude F that hat value of F is required to move the sled up the plane atExplanation / Answer
given
W = m*g = 80 N
mue_s = 0.25
mue_k = 0.15
a) use,
Net force acting on sled parallel to the inclined surface = 0
F_min - m*g*sin(theta) + mue_s*N = 0
F_min = m*g*sin(theta) - mue_s*N
= m*g*sin(theta) - mue_s*m*g*cos(theta)
= 80*sin(20) - 0.25*80*cos(20)
= 8.57 N
b)
use,
Net force acting on sled parallel to the inclined surface = 0
F_min - m*g*sin(theta) - mue_s*N = 0
F_min = m*g*sin(theta) + mue_s*N
= m*g*sin(theta) + mue_s*m*g*cos(theta)
= 80*sin(20) + 0.25*80*cos(20)
= 46.1 N
c)
use,
Net force acting on sled parallel to the inclined surface = 0
F_min - m*g*sin(theta) - fk = 0
F_min = m*g*sin(theta) + mue_k*N
= m*g*sin(theta) + mue_k*m*g*cos(theta)
= 80*sin(20) + 0.15*80*cos(20)
= 38.6 N
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