- #1
trajan22
- 134
- 1
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??
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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