Calculating center of gravity of a pole

[pixietree]

Homework Statement

A horizontal pole is made of 8 pieces, 1 meter apart each. Forces that do not act at the ends are shown in the picture.

The pole does not rotate, and weighs 20N.

a. What is the sum of forces applied on the pole?
b. Where is it applied?
c. What are the forces at the ends of the pole?

Homework Equations

$$ \sum_{k=1}^{n} x_nF_n = 0 $$
Equilibrium point with coordinate k must suffice $$k=\frac {\sum_{k=1}^{n} x_nF_n}{\sum_{k=1}^{n} F_n}$$, where F_i is the i'th force and x_i is its coordinate relative to the leftmost point (taking right to be positive direction), given non-zero sum of forces

The Attempt at a Solution

Let $$\mathbf N_1, \mathbf N_2$$ be the forces applied on the pole at the left, right ends accordingly. I will take the right as shown in the picture to be the positive direction.
Relative to the leftmost point of the pole, the sum of torques is zero because the pole does not rotate.
Hence $$0*N_1-1*10+2*20-7*30+8N_2=0$$, meaning $$\mathbf N_2\ = 22.5\mathbf y\ N $$.
Relative to the rightmost point, the sum of torques is also zero for the same reason.
Hence $$(-8)*N_1+(-7)*(-10)+(-6)*20+(-1)*(-30)=0$$, meaning $$ \mathbf N_1\ = -2.5\mathbf y\ N$$.

And this is where I got stuck - the answer in the book clearly says there is a unique point where the forces may be regarded as being applied there, but the sum of forces is zero if my attempt is right. Probably it was me, so I do not understand what I got wrong.

 

[BvU]

Hello pixietree,

Is this pole massless ?
From you drawing it looks as if the leftmost triangle is a support, not a hinge. So how can it exercise a downward force ?

 

[pixietree]

1- Weird, I triple checked and still didn't write this part. The pole weighs 20N.
2 - This is also another worrysome result I got.

I tried to solve it with 20N force downwards acted somewhere on the pole; my attempt was very cubersome and also failed, so I thought posting this one first is a good idea.

I will re-try again with it and update the post soon.

 

[BvU]

I see you edited post #1 to include the 20 N.

With a mass of (20 N)/g you have a force of 20 N downward at the center of mass, so in the center of the beam.

 

[pixietree]

If I am not mistaken, these are the forces acting on the system:

Suppose the weight acts at distance d from the left.

Relative to the left:
$$1*(-10)+2*20+d*(-20)+7*(-30)+8*N_2=0
-180+8N_2-20d=0$$
$$N_2=22.5+2.5d$$

Relative to the right:
$$(-8)*N_1+(-7)*(-10)+(-6)*20+(d-8)*(-20)+(-1)*(-30)=0$$
$$-8N_1+70-120-20d+160+30=0$$
$$8N_1=140-20d$$
$$N_1=17.5-2.5d$$

Amazingly, according to all other points were forces act, I got tautology (0=0), so they don't add more restrictions.

(Hence, if d is 4 meters, then $$N_1=7.5 \mathbf y N, N_2=32.5 \mathbf y N$$).

Interestingly enough, if I'm right then regardless of d, the sum of forces is 0, leaving me with the original problem!

 

[BvU]

If the pole does not rotate, the sum of torques must be zero.
If the pole does not rotate, the center of mass of the pole does not move
If the center of mass of the pole does not move, the sum of forces (i.e. the net force) must be zero.

I must admit I don't understand the b) question. What does you textbook say about where the resultant of e.g. two forces acts ?

 

[TomHart]

I must admit I don't understand the b) question.

I believe that question a) is asking to find an equivalent force for the three external forces - the -10 N, 20 N, and -30 N forces.
And I believe that question b) is asking to find the location where that equivalent force should be applied to produce the same torque as those three external forces produced. So I guess what you do is sum the torques of the three forces about some point. Then those three forces are replaced with the single resultant force at whatever location produces the same total torque. Depending on the values, that force could be outside of the length of the bar.

For example, let's say there was a 10 meter bar with two forces - one force 10 N downward at the right end of the bar, and the other force 9 N upward 1 meter from the left end of the bar. The resultant force would be -10 + 9 = -1 N (a downward force). The total torque (taken from the left end) of the 2 original forces would be
(9)(1) + (-10)(10) = -91 Nm. So the resultant force of 1 N downward would have to be applied 91 meters to the right of the left edge to produce the same torque as the 2 original forces.

I hope I did that right.

Edit: As I have thought about this after I posted, it just doesn't seem right. Could someone please set me straight. Thank you

 

[haruspex]

question a) is asking to find an equivalent force for the three external forces - the -10 N, 20 N, and -30 N forces.

Yes, or more simply, just the sum of the forces.

question b) is asking to find the location where that equivalent force should be applied to produce the same torque as those three external forces produced. So I guess what you do is sum the torques of the three forces about some point. Then those three forces are replaced with the single resultant force at whatever location produces the same total torque. Depending on the values, that force could be outside of the length of the bar

Yes.

 

[pixietree]

I checked it and the only distance x between consecutive pieces that could sustain equilibrium is, infact, 1 meter.
Weirdly the book says the sum of forces is 30N and acted in distance 0.57 meters (wasn't able to read from where).
The sum of forces is zero. Weird...

 

[haruspex]

I checked it and the only distance x between consecutive pieces that could sustain equilibrium is, infact, 1 meter.
Weirdly the book says the sum of forces is 30N and acted in distance 0.57 meters (wasn't able to read from where).
The sum of forces is zero. Weird...

It is possible you are meant to count the forces from the two end supports as applied forces, but that still makes it 20N, just up instead of down, and it would act in the middle.
Looks like a broken question.