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hyperglitter
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A man drags a 150-lb crate across a floor by pulling on a rope inclined 15 degrees above the horizontal. a) If the coefficient of static friction is 0.50, what tension in the rope is required to start the crate moving?
The answer is supposedly 68 lb.
I start by assuming the normal force = weight, so Fn = 150 lb. The force of friction operating against the direction of motion is Ff = Fn*coefficient of static friction, which is at its maximum.
So 150*.5 = 75 lb = Ff
Now the x component of the force required to move the crate should be equal to the frictional force (Ff). So Cos 15 = 75/Frope, and Frope turns out to be 77.65 lb. This is wrong, but oddly it is 9.8 off from the correct answer (though I realize this is irrelevant since the problem is in lbs). What am I doing wrong? Thanks![Smile :smile: :smile:](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
The answer is supposedly 68 lb.
I start by assuming the normal force = weight, so Fn = 150 lb. The force of friction operating against the direction of motion is Ff = Fn*coefficient of static friction, which is at its maximum.
So 150*.5 = 75 lb = Ff
Now the x component of the force required to move the crate should be equal to the frictional force (Ff). So Cos 15 = 75/Frope, and Frope turns out to be 77.65 lb. This is wrong, but oddly it is 9.8 off from the correct answer (though I realize this is irrelevant since the problem is in lbs). What am I doing wrong? Thanks