A loaded penguin sled weighing 60 N rests on a plane inclined at 20 degreeree to

Question

A loaded penguin sled weighing 60 N rests on a plane inclined at 20 degreeree to the horizontal. Between the sled and the plane the coefficient of static friction is 0.23, and the coefficient of kinetic friction is 0.17. What is the minimum magnitude of the force F, parallel to the plane, that will prevent the sled from slipping down the plane? N What is the minimum magnitude F that will start the sled moving up the plane? N What value of F is required to move the block up the plane at constant velocity? N

Explanation / Answer

a)mgsin(20)-f=F

where f=frictional force,mg=60N

f=60s*cos(20),s=0.23

60sin(20)-60*0.23*cos(20)=F

we F=7.55N

b)as the block is about to move frictiona act in downward direction

60sin(20)+60*s*cos(20)=F

F=33.49N

c)as the block start moving use k

60sin(20)+60*k*cos(20)=F

F=30.10N ask=0.17

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