Finding tension of a rope pulling a crate

1. Oct 15, 2011

Timebomb3750

Did this problem out, but answer doesn't look right.
1. The problem statement, all variables and given/known data
Find the tension of the rope.

M=37.7kg
Coefficient of kinetic friction=.244
Rope is at an angle of 22.2° above the horizontal
Pulled at constant speed, meaning a=0

2. Relevant equations
I figured that...
ƩFx=Tcos∅-μkN=0
ƩFy=N-mg-Tsin∅=0

So, N=mg+Tsin∅

Then I did a simple substitution with the N equation...

Tcos∅-μkmg+Tsin∅=0

When I got T by itself, I got....

T=(μkmg/cos∅+sin∅)

3. The attempt at a solution

Then I just plugged the given numbers in, and I got T=69N (Approximately)

Am I correct? Because I figured the tension of the rope has to be greater than the mg of the crate.

2. Oct 15, 2011

grzz

ƩFy=N-mg-Tsin∅=0

Replace -Tsinangle by +Tsinangle

3. Oct 15, 2011

Timebomb3750

Okay, so that means that my final equation for T changes to...

T=(μk*mg/cos∅-sin∅)

I get an answer of about 164.5 N. It's higher, but is that right. I'm still wondering if T is supposed to be greater than the mg of the crate. If this is correct, then why did I have to change the -Tsin∅, to a positive Tsin∅?

4. Oct 15, 2011

Timebomb3750

bump.

5. Oct 15, 2011

PhanthomJay

When you sum forces in the y direction, N acts up (+), mg acts down (-), and T sin theta acts up (+). That is why grzz corected the signage for Tsin theta.

Beyond that, your algebra is incorrect. There should be a uk in front of the Tsin theta term (note that a(b + c) = ab + ac).

Why do you feel T has to be greater than mg? Consider pulling a heavy crate on a near frictionless surface. The pulling force is much less than mg.

6. Oct 15, 2011

Timebomb3750

Ah, I see now.

Didn't catch the algebra error. Thanks for that.

Meh. I don't know what I was thinking earlier. Was doing this problem early in the morning.

Okay, so now that I corrected my errors, I get a tension of approximately 108 N.

7. Oct 15, 2011

looks good!