Finding tension of a rope pulling a crate

[Timebomb3750]

Did this problem out, but answer doesn't look right.

Homework Statement

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

Homework 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∅)

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.

 

[grzz]

ƩFy=N-mg-Tsin∅=0

Replace -Tsinangle by +Tsinangle

 

[Timebomb3750]

ƩFy=N-mg-Tsin∅=0

Replace -Tsinangle by +Tsinangle

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∅?

 

[Timebomb3750]

bump.

 

[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.

 

[Timebomb3750]

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.

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.

 

[PhanthomJay]

looks good!