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Forces and friction 
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#1
Oct906, 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?? 


#2
Oct906, 10:52 PM

Sci Advisor
HW Helper
P: 3,031

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.



#3
Oct1006, 03:00 PM

P: 155

but since both are variables how could I solve for this



#4
Oct1006, 03:15 PM

Sci Advisor
HW Helper
P: 3,031

Forces and friction



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