Finding Acceleration from Friction Coefficient and Force Applied

[ChrisMihm31]

Homework Statement

To move a large crate across a rough floor, you push on it with a force F at an angle of 21°, below the horizontal, as shown in Figure 6-21. Find the acceleration of the crate, given that the mass of the crate is m = 32 kg, the applied force is 344 N and the coefficient of kinetic friction between the crate and the floor is 0.54.

http://www.webassign.net/walker/06-16alt.gif

Homework Equations

a = F_net/m
F_friction = F_normal * mu_k

The Attempt at a Solution

First I found the frictional force by (.54)(32 kg)(9.81 m/s2) = 169.5168
Now this is where I am stuck. Do you subtract that from the horizontal force? The horizontal force I got was 344Cos(21) = 321.15. Then I do 321.15 - 169.5168 = 151.63 and then divide that by the mass(32kg) and get 4.74

This is obviouley what you aren't supposed to do, so HELP!

 

[Rake-MC]

Sorry but your frictional force is wrong.

Force due to friction is the coefficient of friction multiplied by the normal reaction force.

First you must find the normal reaction force.

 

[ChrisMihm31]

Isn't the normal force just mass x gravity?

 

[Rake-MC]

No, because there is also a vertical component of the force applied by the worker.

Sum up the forces in the y plane to find the normal reaction force.

 

[ChrisMihm31]

so would the normal reaction force be: mg + Fsin(21) ?
probably wrong, but its my best guess

 

[Rake-MC]

That's correct.

 

[ChrisMihm31]

Thanks for your help, I plugged everything in and got the correct answer: 2.66 m/s².