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A loaded penguin sled weighing 87 N rests on a plane inclined at angle θ = 22° to the horizontal (Fig. 6-28). Between the sled and the plane, the coefficient of static friction is 0.26, 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?

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2007-09-22 13:07:27 · 2 answers · asked by James B 1 in Science & Mathematics Physics

2 answers

Oh, the fun we used to have penguin sledding!
The force normal to the plane Fn = 87*cos(theta).
The force parallel to the plane is Fgrav + Ffriction.
Fgrav = 87*sin(theta)
Ffriction = 0.26Fn when stopped, 0.18Fn when moving.
Add Fgrav to the appropriate Ffriction for the answers to b and c.
Subtract Fgrav from the static Ffriction for the answer to a.

2007-09-23 09:54:21 · answer #1 · answered by kirchwey 7 · 0 0

Newton’ 2d regulation states that the acceleration of an merchandise relies upon upon 2 variables - the internet tension appearing upon the object and the mass of the object... The regulation is, F=ma, the place m is the mass of the object that undergoes an acceleration a with the aid of an utilized tension F.

2016-11-06 03:06:15 · answer #2 · answered by datta 4 · 0 0

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