How Does Friction Affect a Penguin Sled on an Inclined Plane?

  • Thread starter Thread starter mmiller9913
  • Start date Start date
  • Tags Tags
    Friction Hill
AI Thread Summary
The discussion focuses on calculating the forces acting on a penguin sled on an inclined plane, specifically addressing static and kinetic friction. The sled weighs 66 N and is inclined at 18°, with coefficients of static and kinetic friction given as 0.27 and 0.18, respectively. To prevent the sled from slipping down, the minimum force required is derived from the balance of forces, including the applied force, static friction, and gravitational components. The calculations reveal that the force needed to start moving the sled up the plane is approximately 17.81 N, while maintaining constant velocity requires about 11.87 N. Understanding the forces at play, including the normal force and friction direction, is crucial for solving the problem effectively.
mmiller9913
Messages
13
Reaction score
0

Homework Statement


A loaded penguin sled weighing 66 N rests on a plane inclined at angle θ = 18° to the horizontal (Fig. 6-23). Between the sled and the plane, the coefficient of static friction is 0.27, and the coefficient of kinetic friction is 0.18. (a) What is the minimum magnitude of the force , 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?


Homework Equations


Fk=uFn
Fsmax=uFn

The Attempt at a Solution


a) I know the acceleration of the sled would be negative but if I set it up like
Fa-Fs=-ma... it can't be solved

b) Fsmax=uFn
=.27(6.73)(9.8)
=17.81

c)Fa-Fk=ma
Fa-Fk=0
Fa=Fk
Fa=uFn
=11.87N
 
Physics news on Phys.org
mmiller9913 said:
a) I know the acceleration of the sled would be negative but if I set it up like
Fa-Fs=-ma... it can't be solved
What forces act on the sled parallel to the plane? There are three, one of them being the applied force that you need to solve for. What are the other two? What must they add to for the sled to not move?
 
I don't see the three forces that act on it...the applied force, the static friction force, and?
 
mmiller9913 said:
I don't see the three forces that act on it...the applied force, the static friction force, and?
What force is always there, acting on any mass? (What force acts to make it slide down the hill?)
 
okay so ...
Fa-Fs-Fgcostheta=-ma

I don't see how this can be solved though
 
you need to sum the forces in the Y as well, so you can find the normal force
 
mmiller9913 said:
okay so ...
Fa-Fs-Fgcostheta=-ma

I don't see how this can be solved though
What direction does friction act? What does the friction force equal? What's the 'cos theta' business? What's the acceleration? (Remember--you are preventing it from sliding.)
 
Back
Top