forces and friction


by trajan22
Tags: forces, friction
trajan22
trajan22 is offline
#1
Oct9-06, 08:40 PM
P: 155
A factory worker pushes a 28.8 kg crate a distance of 4.25 m along a level floor at constant velocity by pushing downward at an angle of 28.8 degrees below the horizontal. The coefficient of kinetic friction between the crate and floor is 0.250 .
What magnitude of force must the worker apply to move the crate at constant velocity?

what i did is take it that the force pushing the crate must be equal to the force of friction since there is no acceleration.
F(friction)=mg mu_k therefore 28.8*9.8*.25
this is the force in the x direction therefore
tan(28.8)*70.56(force of friction)=magnitude of the force in y direction i then use c^2=a^2+b^2 to find total force which i get to be 80.51N
however i am told the correct answer is 93.3N
where have i gone wrong??
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OlderDan
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#2
Oct9-06, 10:52 PM
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The worker is not pushing the crate with a horizontal force. By pushing downward, the worker is increasing the normal force required to support the crate and increasing the friction. You need to account for this extra force.
trajan22
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#3
Oct10-06, 03:00 PM
P: 155
but since both are variables how could I solve for this

OlderDan
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#4
Oct10-06, 03:15 PM
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P: 3,033

forces and friction


Quote Quote by trajan22
but since both are variables how could I solve for this
You know the direction of the force. The only unknown you need to find is the magnitude of the force. Look at the free body diagram and account for all of the forces. The normal force will now depend on the magnitude of the applied force, but it can be expressed in terms of given quantities and the unknown force. Your horizontal equation will then have only one unknown.


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