Calculating Tension in Arms of 56.2kg HS Student

[Cole07]

This is my problem:
1. A 56.2-kg high-school student hangs from an overhead bar with both hands. What is the tension in each arm if the bar is gripped with both arms raised vertically overhead?
2. What is the tension in each arm when the arms make an angle of 35.00 o with respect to the vertical?
I have two equations:
A.) sum of all forces=ma
B.) sum of all forces=0

i tried:
sum of all forces=56.2*9.81
but this was wrong

then for the second question i tried:
sum of all forces=56.2cos(35.00)
i found this equation after research and don't know too much about it.

I don't understand how to plug this into these equations or if the are the right ones. Could someone lead me in the right direction. any help would be greatly appreciated.

 

[Dorothy Weglend]

This is my problem:
1. A 56.2-kg high-school student hangs from an overhead bar with both hands. What is the tension in each arm if the bar is gripped with both arms raised vertically overhead?
2. What is the tension in each arm when the arms make an angle of 35.00 o with respect to the vertical?
I have two equations:
A.) sum of all forces=ma
B.) sum of all forces=0

i tried:
sum of all forces=56.2*9.81
but this was wrong

then for the second question i tried:
sum of all forces=56.2cos(35.00)
i found this equation after research and don't know too much about it.

I don't understand how to plug this into these equations or if the are the right ones. Could someone lead me in the right direction. any help would be greatly appreciated.

Sounds to me like you need to draw a free body diagram, and that would sort out the confusion in short order.

Once you have this, translating it into mathematics and solving it is not difficult.

Dorothy

 

[ponjavic]

What do you mean by sum of all forces = ...

That would literally mean that there would be no equilibrium and the student would be falling or flying or something.

First thing you do draw a FBD (see pic)

Now for equilibrium the two forces on the arms have to equal the force (weight) on the bar (coming from the person).

Now you should be able to work out the force on one arm!

 

[Cole07]

Ok i drew a free body diagram and i see
W=56.2
T1=?
T2=?
and i know that T1 and T2 are the same

 

[Cole07]

"Now for equilibrium the two forces on the arms have to equal the force (weight) on the bar (coming from the person)."

would this be 551.322
(56.2*9.81)

 

[Dorothy Weglend]

Ok i drew a free body diagram and i see
W=56.2
T1=?
T2=?
and i know that T1 and T2 are the same

Good for you. Except W = mg, not m, right?

So T1 = T2, so just call it T. Call the vectors (T) pointing up positive. Add them together, subtract the vector going down, and then decide what they are all equal to.

Then solve for T, and you are done.

The same approach will work for part B. It is slightly more complicated, but only slightly.

Dorothy

 

[ponjavic]

Ok i drew a free body diagram and i see
W=56.2
T1=?
T2=?
and i know that T1 and T2 are the same

For equilibrium the sum of all the forces is equal to zero

IE

W + T1 + T2 = 0, and as you said T1 = T2 Solve this?

Ignore my FBD it was rubbish I forgot the weight of the person

I'll just leave this one to Dorothy ^^

 

[Cole07]

"So T1 = T2, so just call it T. Call the vectors (T) pointing up positive. Add them together, subtract the vector going down, and then decide what they are all equal to."
To find what T is would you do 551.322+T=0 ?

 

[ponjavic]

"So T1 = T2, so just call it T. Call the vectors (T) pointing up positive. Add them together, subtract the vector going down, and then decide what they are all equal to."
To find what T is would you do 551.322+T=0 ?

Is there just one T? As T1=T2

 

[Cole07]

no there is two and they equal the same number. but i don't understand how to get the T1 or T2 i have the equation (sum of all forces)=0
and the sum of all force is sigmaF, is this not the equation i use?

 

[Cole07]

W + T1 + T2 = 0
can you leave one T out since they are the same and it just be W+T=0 ?

 

[Dorothy Weglend]

Hi Cole,

Your FBD diagram should have two vectors pointing up, representing the forces of the arms. And one vector pointing down, that represents the force of gravity, the weight.

So if the two vectors are pointing up (call that positive, as I suggested), and the weight vector is pointing downwards, then you have:

Sumof Forces in Y: 2T - mg = ?

I left the ? for you to fill in...

Dorothy

 

[Dorothy Weglend]

W + T1 + T2 = 0
can you leave one T out since they are the same and it just be W+T=0 ?

You can't leave out one T, because that doesn't represent the physics of the situation. That would be the case if the person were hanging by just one arm, and the force on that arm would be much greater than if he is hanging by two arms.

Dorothy

 

[ponjavic]

W + T1 + T2 = 0
can you leave one T out since they are the same and it just be W+T=0 ?

No you cannot, it is like this: If T1=T2 then:

W+T1+T1=0 thus W+2T1=0 and mind the vectors as dorothy said IE

W-T1-T2=0 W-T1-T1=0 W-2*T1=0

 

[Cole07]

oh this makes a lot more since i used your formula 2T-mg=0

2T-56.2(9.81)=0

2T/2=551.322/2

T= 275.661

 

[ponjavic]

oh this makes a lot more since i used your formula 2T-mg=0

2T-56.2(9.81)=0

2T/2=551.322/2

T= 275.661

So just by thinking about it logically you should understand that the force from the weight gets split up between each arm!

 

[Cole07]

I still don't under stand what to do when i have an angle though would i use the formula
sum of all forces=56.2*9.81cos(35.00)

 

[Cole07]

So just by thinking about it logically you should understand that the force from the weight gets split up between each arm!

yes i did understand this part.

 

[Cole07]

Is the vertical the person?

 

[Dorothy Weglend]

I still don't under stand what to do when i have an angle though would i use the formula
sum of all forces=56.2*9.81cos(35.00)

Stop searching for formulas

When you have forces at an angle like this, you need to break them up into the component vectors.

If the arms are at an angle to the verticle, then part of the force on the arm is going straight up and down, and part is going left and right. These are the y and x components of the vectors. You can use trigonometry to get these.

Then you balance these forces in the Y and X direction, just as you did for the Y direction in part A.

Isn't this covered in your textbook, or by your teacher? Just curious.

Dorothy

 

[Cole07]

I can't find anything about tension in my textbook anywhere everything else we are studing is in there. when you say y and x components can i make a chart where i use cos for x and sin for y we did something like this in the last chapter for velocity?

 

[ponjavic]

You need to sit down with your teacher and have a good luck at free body diagrams and vectors

 

[Cole07]

i think i do see where the angle is now is formed when the bar tilts it is formed by the bars new location and the bars old location?

 

[Dorothy Weglend]

i think i do see where the angle is now is formed when the bar tilts it is formed by the bars new location and the bars old location?

The bar tilts? I thought the position of the arms changed from straight up and down, to more spread, so that each arm and the vertical makes an angle of 35 degrees.

But I agree, you need to sit down with your teacher and have a chat about this stuff. For now, I have uploaded a freebody diagram of the new situation. Your job is to understand it, label the forces, and create your own formulas, one for the X forces, and one for the Y forces.

Dorothy

 

[Cole07]

no sorry it doesn't tilt i misread the question