A loaded penguin sled weighing 89 N rests on a plane inclined at angle = 24° to

Question

A loaded penguin sled weighing 89 N rests on a plane inclined at angle = 24° to the horizontal (see the figure). Between the sled and the plane, the coefficient of static friction is 0.24, and the coefficient of kinetic friction is 0.18. (a) What is the minimum 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've seen how people have answered the question on here, but I don't understand where all the formulas are coming from. My professor only gave the formulas f (friction) = (constant coefficient) (Normal vector) and said that was the only formula we needed for these kinds of problems. If the normal vector has the same magnitude as W but in a different direction, where is Wcos(theta) and all these values coming from? Is there a general way of expressing these kinds of scenarios with different formulas that will always work with these problems?

Explanation / Answer

a) The sled at rest yields force parallel to the incline as
F+89*cos(24)*0.24-89*sin(24)=0
solve for F
F=16.68 N

b)
F-89*cos(24)*0.24-80*sin(24)=0
solve for F
F=55.71 N

c)
F-89*cos(24)*0.18-89*sin(24)=0
solve for F
F=50.83 N

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.