Calculating Mass of 4.2m Post Suspended by 1,700N Cable

[brentwoodbc]

Homework Statement

A 4.2 m long uniform post is suspended by a cable having a tension of 1 700 N.

Homework Equations

What is the mass of this post?

The Attempt at a Solution

I figure there are 2 forces acting in a clockwise direction (the mass of the log on each side of the rope. And that each of those masses should be multiplied by sin60 not sure about the tension though. I figure that the tension of the rope is acting in a counter clockwise direction, and that all torques are = 0.)

the answer is 530kg.

 

[LowlyPion]

Is that your answer? Or the book's?

For instance I appreciate the hints, but I think I know how to work it.

What are you stuck on is what I'm asking?

 

[brentwoodbc]

Is that your answer? Or the book's?

For instance I appreciate the hints, but I think I know how to work it.

What are you stuck on is what I'm asking?

No I don't know how to solve it. The booklet I have has the answer of 530 kg but I don't understand how to get to that solution.

I gave the info on how I think I should do it but I am stuck/confused.

 

[LowlyPion]

No I don't know how to solve it. The booklet I have has the answer of 530 kg but I don't understand how to get to that solution.

I gave the info on how I think I should do it but I am stuck/confused.

OK. If the sum of the torques is 0, what are the torques about the pivot?

 

[brentwoodbc]

thats what's confusing me, What about the tension of the rope? Is does that have torque?

I think its like
$$\tau$$1+$$\tau$$2-Tr=0
m(9.8)sin60(2.1)+m(9.8)sin60(.5)-(1700)(4.2)
17.8m+4.24m=7140
22.04m=7140

22.04m=7140
$$\overline{22.04=22.04}$$

m=323.96kg...thats a stab at it.

 

[LowlyPion]

thats what's confusing me, What about the tension of the rope? Is does that have torque?

What about the tension in the rope? Isn't that a force? Isn't it acting over a moment arm about the pivot? And how far away from the pivot is it acting ? So what is the torque from the tension in the rope?

 

[brentwoodbc]

What about the tension in the rope? Isn't that a force? Isn't it acting over a moment arm about the pivot? And how far away from the pivot is it acting ? So what is the torque from the tension in the rope?

thats why I did the force of tension 1700xdistance from pivot (4.2)

But I assume that you need to do cosine60 or something?

 

[LowlyPion]

thats why I did the force of tension 1700xdistance from pivot (4.2)

But I assume that you need to do cosine60 or something?

The log is 4.2m. How far from the pivot is the Tension acting?

 

[brentwoodbc]

darn I read it wrong so 3.2m

 

[brentwoodbc]

that gives me an even lower number though (246kg)

 

[LowlyPion]

Now what other Torque is that acting against?

 

[brentwoodbc]

Now what other Torque is that acting against?

wouldnt that make the weight even less?

I can't think of any others other than maybe the ground or the angle of the log?

 

[LowlyPion]

wouldnt that make the weight even less?

I can't think of any others other than maybe the ground or the angle of the log?

Where is the center of mass acting?

 

[brentwoodbc]

Where is the center of mass acting?

thats one of the things that's confusing me is it 2.1 metres?
or something like 2.1(sin60)

 

[LowlyPion]

thats one of the things that's confusing me is it 2.1 metres?
or something like 2.1(sin60)

It is at 2.1m along the log. But what distance is the projection of m*g acting from the pivot?

Draw a line straight down. That distance from the pivot along the ground is the projection for a ⊥ moment the weight is acting through.

 

[sArGe99]

See how much weight would provide torque in this case?
Its not all of it.

 

[brentwoodbc]

isnt it just 2 torques the log clockwise and the rope counterclockwise

mgsinθr=1700r
m(9.8)sin60(2.1)=1700(3.2)
17.8m=5440
m=305kg?

 

[sArGe99]

Very close. Resolve the vectors again and see which component counters torque due to tension in the thread.

 

[brentwoodbc]

so its cosine. right? thanks!

 

[sArGe99]

If it wasn't the sine function, it had to be the cosine function obviously. But please understand why its the cosine function. And try more problems related to resolution of vectors to get that idea clear.

 

[brentwoodbc]

If it wasn't the sine function, it had to be the cosine function obviously. But please understand why its the cosine function. And try more problems related to resolution of vectors to get that idea clear.

I don't really understand why its cosine, I guess its because the forces need to be parrall, I still don't get it though.

 

[LowlyPion]

I don't really understand why its cosine, I guess its because the forces need to be parrall, I still don't get it though.

It's cosine in this case because that is the distance through which the CoM acts ⊥ to the pivot. When figuring moments and angular momentum etc this is the important distance. The rest is trig to determine the point at which it acts ⊥ to an arm, any arm, to the pivot.

The vector form of the Torque equation is T = F X r , the vector cross product, and this relationship relies on the ⊥ action about a point. You may not need to know that now, but it is on the road ahead if you get to it.