Calculating Frictional Forces on a Loaded Penguin Sled on an Inclined Plane

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A loaded penguin sled weighing 80N is on a 20-degree inclined plane, with static and kinetic friction coefficients of 0.25 and 0.15, respectively. To prevent the sled from slipping down, a force of 196N is required, while a minimum force of 118N is needed to start moving it up the plane. The calculations involve applying the second law of dynamics and incorporating the friction coefficients. For maintaining constant velocity while moving up the plane, further clarification is needed, as the user struggles with deriving the correct equation. The discussion highlights the importance of understanding frictional forces in inclined planes.
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A loaded penguin sled weighing 80N rests on a plane inclined at angle = 20 degrees to the horizontal. Between the sled and the plane, the coefficient of static friction is 0.25, and the coefficient of kinectic friction is 0.15. (a) What is the least magnitude of the force F, parallel to the plane, that will prevent the sled from slipping down the plane? (b) What is the minimum magnitude F that will start the sled moving up the plane? (c) What value of F is required to move the sled up the plane at constant velocity?...i am totally lost on the parallel part and the velocity, iam not quite understanding how to derive an equation for it...need help please... :confused:
 
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Did u solve point a)...?It used the static friction coeff.As for the last 2 points,the key is the same:apply the II-nd law of dynamics...This time you'll be using the kinetic friction coeff...

Daniel.
 
for a.) 196N b.) 118N c.) ...? for part a : Uk(Mg+sin20)= .25(80N*9.8m/s2 + sin20) = 196N , for part b: Us(Mg+cos20)= .15(80N*9.8m/s2+ cos20)= 118N...and for part c i don't know, i tryed using the the equation for part b, and i divided by 118N i got "0" i know its wrong...need help BIG TIME..
 

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