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- Thread starter Shah 72
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- Jan 30, 2018

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The friction force, opposing the motion, is the coefficient of friction times the downward force In this case the downward force is the weight of the box, 6g N (g is the acceleration due to gravity, 9.81 meters per second squared), plus the downward force applied by the boy, Calling that force, F, that is a total downward force, 6g+ F newtons so the friction force is 0.35(6g+ F) Newtons. Since the box " is on the point of slipping but remains at rest" that must be equal to the applied force 25 Newtons.

Solve 0.35(6g+ F)= 25 for F.

Solve 0.35(6g+ F)= 25 for F.

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Thank you so much!!!The friction force, opposing the motion, is the coefficient of friction times the downward force In this case the downward force is the weight of the box, 6g N (g is the acceleration due to gravity, 9.81 meters per second squared), plus the downward force applied by the boy, Calling that force, F, that is a total downward force, 6g+ F newtons so the friction force is 0.35(6g+ F) Newtons. Since the box " is on the point of slipping but remains at rest" that must be equal to the applied force 25 Newtons.

Solve 0.35(6g+ F)= 25 for F.