Solving the Net Force on a Wooden Box

[brentwoodbc]

Homework Statement

A 250 N force is applied at an angle of 32° above the horizontal to a 96 kg wooden box causing
it to slide along a floor as shown.

The coefficient of friction between the floor and the box is 0.18. What is the magnitude of the net
force on the wooden box?

The Attempt at a Solution

ff=mu times N
ff =0.18 times 9.8 times 96
For normal force would you find the y component ie. 9.8 times 96 times sin32 or do you just do N=W?
ff=170N

fapp=250cos32
fapp=212

Fnet=fapp-ff
fnet=212-170=43N?
and if I do N as the y component "because some of the mass is being lifted" I get the wrong answer too. 122N
it is supposed to be 67N?

 

[dx]

To calculate the normal force, you need to consider the y component of the applied force too.

 

[brentwoodbc]

To calculate the normal force, you need to consider the y component of the applied force too.

I know, I just can't remember if for ff you multiply N times sin 32, I don't remember doing that. so the y component of the applied force is subtracted from the normal force?

 

[dx]

It is subtracted from the weight. N = W - F sin(32°). Tell me if you don't see why.After you get the normal force, the friction force is just f = µN.

 

[brentwoodbc]

It is subtracted from the weight. N = W - F sin(32°). Tell me if you don't see why.


After you get the normal force, the friction force is just f = µN.

Alright I got it, Its been a while and I am just doing some review. "also I was using an online calculator that was giving me some odd answers."
Thank you a lot.

 

[dx]

No problem.