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 .(adsbygoogle = window.adsbygoogle || []).push({});

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??

**Physics Forums | Science Articles, Homework Help, Discussion**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Forces and friction of pushed crate

**Physics Forums | Science Articles, Homework Help, Discussion**