Hanging a deer from a pole - find tension

[ldbaseball16]

Homework Statement

Ricky Bobby hangs a deer from a uniform horizontal pole as shown below. The Deer's mass is 57.6 kg, the pole has a mass of 12.1kg, and the length of the pole is 5.36m. The Deer is hung 1.92m from the point where the pole is attached to the vertical beam, and the angle of the support wire at the end of the pole 21.3 degrees. What is the tension in the wire? what is the direction and the magnitude of the force of the vertical support against the pole at the point where they meet?

Homework Equations

T=<T<145degrees>
W=<57.6<270degrees>
Weight of pole=<1.92<270degrees>
force on pole=<F<21.3>

The Attempt at a Solution

<T<145>+<57.6<270>+<1.92<270>+<F<21.3>=0 I don't know if I am doing this right can you help me please

 

[tiny-tim]

Homework Statement

Ricky Bobby hangs a deer from a uniform horizontal pole as shown below. The Deer's mass is 57.6 kg, the pole has a mass of 12.1kg, and the length of the pole is 5.36m. The Deer is hung 1.92m from the point where the pole is attached to the vertical beam, and the angle of the support wire at the end of the pole 21.3 degrees. What is the tension in the wire? what is the direction and the magnitude of the force of the vertical support against the pole at the point where they meet?
…
T=<T<145degrees>
W=<57.6<270degrees>
Weight of pole=<1.92<270degrees>
force on pole=<F<21.3>
…
<T<145>+<57.6<270>+<1.92<270>+<F<21.3>=0 I don't know if I am doing this right can you help me please

Hi ldbaseball16!

Yes, that's certainly right in principle …

you've correctly applied good ol' Newton's second law for the forces on the end of the pole …

but your (r,θ) notation is useless for adding the vectors … you'll need to convert to (x,y) …

and you'll also need to take moments about the other end of the pole.

 

[ldbaseball16]

yea but how does 5.36m of the pole and 12.1 kgs tie into the equation? that's what throws me off

 

[tiny-tim]

yea but how does 5.36m of the pole and 12.1 kgs tie into the equation? that's what throws me off

Hi ldbaseball16!

The 5.36m comes in when you take moments about the other end of the pole.

 

[ldbaseball16]

i don't understand you mean like this ? <T<145>+<1.92<270>+<12.1<180>+<5.36<180>+<57.6<270>+<F<21.3>=0 and then convert to (x,y)?? right

 

[tiny-tim]

i don't understand you mean like this ? <T<145>+<1.92<270>+<12.1<180>+<5.36<180>+<57.6<270>+<F<21.3>=0 and then convert to (x,y)?? right

wow, that was three months ago …

that deer must be smelling pretty niffy by now

sorry, but I can't follow your equation …

can you please convert it to x and y first, and then ask me?

 

[ldbaseball16]

Tcos(145)+1.92cos(270)+12.1cos(180)+5.36cos(180)+57.6cos(270)+Fcos(21.3)=0

Tsin(145)+1.92sin(270)+12.1sin(180)+5.36sin(180)+57.6sin(270)+Fsin(21.3)=0

 

[tiny-tim]

Tcos(145)+1.92cos(270)+12.1cos(180)+5.36cos(180)+57.6cos(270)+Fcos(21.3)=0

Tsin(145)+1.92sin(270)+12.1sin(180)+5.36sin(180)+57.6sin(270)+Fsin(21.3)=0

uhh?

where did 145º come from?

and please write it out properly, with all angles < 90º, and putting cos or sin = 0 or 1 where appropriate

 

[ldbaseball16]

Tcos<145>+1.92cos<270>+12.1cos<180>+5.36cos<180>+5 7.6cos<270>+Fcos<21.3>=0

Tsin<145>+1.92sin<270>+12.1sin<180>+5.36sin<180>+5 7.6sin<270>+Fsin<21.3>=0


the <145> is the angle of the missing tension of the wire

 

[tiny-tim]

the <145> is the angle of the missing tension of the wire

no it isn't …

Ricky Bobby hangs a deer from a uniform horizontal pole as shown below. The Deer's mass is 57.6 kg, the pole has a mass of 12.1kg, and the length of the pole is 5.36m. The Deer is hung 1.92m from the point where the pole is attached to the vertical beam, and the angle of the support wire at the end of the pole 21.3 degrees. What is the tension in the wire? what is the direction and the magnitude of the force of the vertical support against the pole at the point where they meet?

and what is sin180º, for example?

 

[ldbaseball16]

Tcos(145)+1.92cos(270)+12.1cos(180)+5.36cos(180)+5 7.6cos(270)+Fcos(21.3)=0

Tsin(145)+1.92sin(270)+12.1sin(180)+5.36sin(180)+5 7.6sin(270)+Fsin(21.3)=0

T+0-12.1-5.36+0+F.9316912276=0

T-1.92+0+0-57.6+F.3632512305=0

so it just stays as T and no <145>?