- #1
Timebomb3750
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Did this problem out, but answer doesn't look right.
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
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∅)
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.
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.