Calculate the tension in the string

[ssj]

1. Calculate the tension in the string T in each of the following situations.
In all calculations ignore the mass of the beam.

http://img146.imageshack.us/img146/7291/43422126ru2.th.jpg


2. Moment=F X Perpendicular Distance and Basic Trigonometry



3.I have tried 200N X 0.5m=100 100/Cos30

Help is much apperchated.

 

[Doc Al]

I have tried 200N X 0.5m=100 100/Cos30

I understand the left side of this equation: the clockwise torque due to the 200 N force. The right side should be the counter-clockwise torque due to the tension of the cable.

Correct this equation, rewriting the right hand side in terms of T (tension). Then solve for T.

 

[ssj]

I understand the left side of this equation: the clockwise torque due to the 200 N force. The right side should be the counter-clockwise torque due to the tension of the cable.

Correct this equation, rewriting the right hand side in terms of T (tension). Then solve for T.

I think i have found the solution by calculating the "opp" tan30 X 0.5 = 0.2886 from there onward we do 200 X 0.5= 100/0.2886 =346.5/cos 30 = 400 N

However now I am stuck on an similar equation of http://img374.imageshack.us/img374/3332/22917237wh9.th.jpg

 

[Doc Al]

I think i have found the solution by calculating the "opp" tan30 X 0.5 = 0.2886 from there onward we do 200 X 0.5= 100/0.2886 =346.5/cos 30 = 400 N

Don't really understand what you did there with tangents and cosines. All you need is:
0.5*200 = 0.5*T*sin(30)
which gives you T = 400.

However now I am stuck on an similar equation of...

Solve it in the same way. The only difference is the moment arm associated with the 200 N force.

I strongly suggest you adopt a systematic approach, like the one I used.

 

[ssj]

But why would you use sine30 wouldn't that work out the "opp"?. Due to the fact the T is on the hypoteanus.

 

[Doc Al]

But why would you use sine30 wouldn't that work out the "opp"?. Due to the fact the T is on the hypoteanus.

You want the component of T perpendicular to the beam, which in this case means T\sin {30}.

 

[ssj]

But why would T need to become perpendicular ? Also for question 2 how am I ment to work it out without the given length ?

 

[Doc Al]

But why would T need to become perpendicular ?

Because you are finding the torque (moment) it generates about the pivot point. Do you understand the definition of moment?

 

[ssj]

Because you are finding the torque (moment) it generates about the pivot point. Do you understand the definition of moment?

The moment is the turning force right ?

 

[Doc Al]

The moment is the turning force right ?

You could call it that, yes.

 

[ssj]

Im stuck on the second question I posted because they haven't stated the length what do I do ?

 

[Doc Al]

Im stuck on the second question I posted because they haven't stated the length what do I do ?

You are given all the lengths you need. (The force is at the midpoint of the beam.)

 

[ssj]

So the length is the mid-point between 0 and 0.5 ? meaning its 0.25

 

[Doc Al]

So the length is the mid-point between 0 and 0.5 ? meaning its 0.25

That's correct. But realize that you don't need the length of the beam: You can just use L or L/2. (The actual length doesn't matter.)

 

[ssj]

But doesn't the formula state Motion=Force * perpendicular distance ? This would mean the formula is 200*L = L*T*sin30 correct ?

 

[Doc Al]

But doesn't the formula state Motion=Force * perpendicular distance ? This would mean the formula is 200*L = L*T*sin30 correct ?

Sure. Notice that the length cancels from both sides of that equation.

 

[ssj]

So your left with 200/sin30= T meaning 400= T ?

 

[Doc Al]

So your left with 200/sin30= T meaning 400= T ?

Right, as I stated in post #4. (This is for your first problem.)

 

[ssj]

This is for the second problem though, wouldn't the Motion be different due to the force being applyed to the mid-point? I seem to of gotten T=400 in my previous post.

 

[Doc Al]

This is for the second problem though, wouldn't the Motion be different due to the force being applyed to the mid-point? I seem to of gotten T=400 in my previous post.

These calculations apply to the first problem.

For the second problem, the method is exactly the same: Set clockwise and counter-clockwise torques equal. Of course the torque exerted by the 200 N force is different in the second problem, since it's applied at the midpoint instead of at the end.

 

[ssj]

Right, so I am correct when 200*L=L*T*Sin30 the "L"'s cancel out leaving 200=T*Sin30
We then Divide sin30 on each side leaving 200/sin30=T meaning 400=T is this correct ?

 

[Doc Al]

Right, so I am correct when 200*L=L*T*Sin30 the "L"'s cancel out leaving 200=T*Sin30
We then Divide sin30 on each side leaving 200/sin30=T meaning 400=T is this correct ?

You are correct for the first problem, but not if you are talking about the second problem. (In the second problem, the moment arm for the 200 N force is no longer L.)

 

[ssj]

What is it then in the second problem?

 

[Doc Al]

What is it then in the second problem?

You tell me. That force is applied at the midpoint of the beam, so what do you think?

 

[ssj]

I think that it becomes 200*0.25=0.25*t*sin30 is this equation correct ? 0.25 because 0.5/2= 0.25 also now the center of gravity has changed.

 

[Doc Al]

I think that it becomes 200*0.25=0.25*t*sin30 is this equation correct ? 0.25 because 0.5/2= 0.25 also now the center of gravity has changed.

Not exactly. The only thing that changed is the point of application of the 200 N force; the other force (the tension) is still applied at the same place.

 

[ssj]

Not exactly. The only thing that changed is the point of application of the 200 N force; the other force (the tension) is still applied at the same place.

Wouldnt this mean that the pivot is slightly more changing the angle as the force is closer to the motion.

 

[Doc Al]

The pivot isn't changing, but the distance to the pivot point is. (At least for the 200 N force.)

 

[ssj]

Alright so the perpendicular distance is changing its now at the mid point the half of 0.5 is 0.25 and due to the fact that the force has moved half way I assume the distance is now half as well thus making it 0.25.

 

[Doc Al]

Right. If the beam has length L, then that 200 N force is a distance of L/2 from the pivot.