1. Nov 11, 2013

### hbk69

Hi, i struggle to understand the topic which the following two questions come from? is it centre of gravity/equilibrium of forces? and what are the fundamental concepts/formula's involved in-order to solve the problems as i struggled to picture the problems. Thank you very much for any help. This is the first time i have approached these type of questions so wanted help which would help me answer similar problems in the future

1) Two men are carrying a 2m long 10 kg pole which from which is hung a 80 kg deer that they have shot. How far from the end of the pole should the deer be hung so that the weaker of the two men only supports half the load of the stronger?

For this question, this was my attempt:

X= distance from how far the pole should

X= (2)(10) + (2)(80) / 10 + 90 = 2 cm

When they are asking for how far the deer is supposed to be hung so the weaker man supports half the load of the stronger what are they exactly asking me as i did not seem to understand? is it the centre of gravity? and how can i know who would be the weaker of the two? or are they just saying the weight of the deer is to be split between how much load each man would have to lift, so you'd have one man on one end and one man on the other with the deer in the middle so the questions asks to find the COG? if so how? but i am not sure i have understood the question correctly.

2) A uniform wooden beam 6 m long weighing 80 kg is lying horizontally on the floor. A builder raises one end until the beam is inclined at 300 to the horizontal and holds it in this position until his mate arrives to lift the other end. What is the magnitude of the force he exerts?

And for this question i was trying to think of the forces acting on the beam when the builder raises it, so you have mg=80kg, and then Fsinθ the horizontal component and Fcosθ the vertical component and then i was trying to think of how the resultant would be the force the builder uses to raise the beam and how 6m is supposed to be used

Thank you very much again, any help appreciated

2. Nov 11, 2013

### voko

Yes, this is about the center of gravity and equilibrium. What you have not mentioned, and what is relevant here, is moment of force (torque). Do you know what that is? Do you know of its importance in equilibrium?

3. Nov 12, 2013

### hbk69

I know that torque is a turning force but not of its importance too much, i understand that it is used to calculate turning force when objects are being rotated but in this case i can't see where the rotation would be

4. Nov 12, 2013

### voko

Equilibrium is an ABSENCE of rotation. See?

5. Nov 12, 2013

### hbk69

Equilibrium is where all the net force is zero and forces are balanced so you are saying that is the point where rotation is absent, but how can i identify it in the problem the rotation bit

6. Nov 12, 2013

### voko

The rotation is caused by torque. If there non-zero net torque on a system, it will rotate. So what condition do you obtain for equilibrium in terms of torque?

7. Nov 12, 2013

### hbk69

There is a pivot point on the body from which it rotates when a force is applied? and at the pivot the net force is zero? not sure

8. Nov 12, 2013

### voko

The net force must be zero in equilibrium. For example, if a massive ball rests on a hard horizontal surface, its weight is mg, and the normal force is also mg in magnitude, but opposite in direction; the net force is zero.

But zero net force is not sufficient for equilibrium. For example, if you try to suspend horizontally a massive bar at a single point, which is not its center of mass, then no matter what force you apply at that single point, the bar will not be in equilibrium.

9. Nov 14, 2013

### hbk69

How could i calculate the answers to the questions please? I don't know where to begin due to the problems stated in the OP, i see that the net force must be zero in equilibrium though

10. Nov 14, 2013

For an object to be in equilibrium,you need two things:
1-Net force is zero
2-Sum of the clockwise moment about any point is equal to the sum of the anticlockwise moment about that point

Moment=F x perpendicular distance
Try to draw a Free Body diagram.

Look at Diagram A,Here Net force is zero,but 2 doesnt apply-Not equilibrium
Diagram B,again net force is zero,2 doesn't apply-Not equilibrium
Diagram C,Net force is zero,2 apply-Equilibrium

If Net force is not zero,It will be accelerating
If Sum of the clockwise moment about any point is not equal to the sum of the anticlockwise moment about that point,It will be rotating

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11. Nov 14, 2013

### hbk69

so for problem 1 its force x perpendicular distance, and in-order to find the distance of how far the deer is i rearrange the Moment=Fd formula where d is the distance of how the deer needs to be hung, d=moment/F, the force acting downwards is 80+10 but not sure how to get the moment

12. Nov 15, 2013

### voko

The pole is massive; this means the force of its weight is applied at its center of mass.

Then there is the force due to the weight of the deer; it is applied at an unknown position.

Finally, there are two forces applied to the pole by the hunters at the two ends of the pole.

All of these forces and their moments must sum to zero so that the pole be in equilibrium.

13. Nov 15, 2013

### hbk69

How can i calculate the forces which the men are applying at each end of the pole and the unknown position of the deer

14. Nov 15, 2013

### voko

The forces are unknown. But you know that one is twice the other; and that together they must support the weight of the pole and the deer. That gives you two equations from which you can determine their magnitudes.

15. Nov 15, 2013

Object is in equilibrium

Firstly,find the two forces,weaker man gives F/2 and stronger man gives F.Together these two forces support the weight of the pole and deer.You can find the force this way.-Net force is zero

To find the distance,you have to make use of the "Sum of clockwise moment of any point is equal to sum of anticlockwise about that point.
The clockwise moment of the point, weaker man(A), should be equal to the anti clock wise moment about that point.The force given by the stronger man provides the anticlockwise moment.Clockwise moment is provided by the weight of pole and deer.
τ=F x d

Look at the free body diagram.Once you find the force F,You can find the anticlockwise moment.Using the weight of the pole and deer,clockwise moment
Then you can solve for d

Even if you find the moment at point,stronger man,You will get the same answer because sum of clockwise moment about any point is equal to the sum of anticlockwise moment about that point

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16. Nov 29, 2013

### hbk69

Hi, i have been doing some work on equilibirum basics etc to help me approach this problem so i would understand better when you help me.

In this problem the net force is zero as everything is in equilibirum, so that means the sum of all the forces would be equal zero, or the upwards forces = the downwards forces.

I see that one of the men is half of the strength of the other so one will have a force of F1 while the other F1/2.

But they have not mentioned where the deer is, do we assume it is at the centre of the pole and add its weight to the weight of the pole?

If we add all the forces we have F1 + F2/2 = 9.8*10 + 80*9.8

If we have the weaker man on the left end and take that as the pivot and the stronger on the right then apply the sum of clockwise = anti-clockwise moments we get:

F1*2m = (9.8*10)*1 + (80*9.8)d where the d is the distrance of the deer from the pivot, where d is the "How far from the end of the pole should the deer be hung so that the weaker of the two men only supports half the load of the stronger"? is that what they are looking for when they are asking for how far the deer is hung?

I know i am wrong in my working out but i feel i am in a better position then before when it comes to my understanding so could you please help me? i really want to be able to find the solution and help my understanding

Any help thanks

17. Nov 29, 2013

### hbk69

Also don't know how to work out any of the positions on the pole but the above post is the best i could do at the moment

18. Nov 29, 2013

### hbk69

Hi, again i have managed work out the first question so ignore my previous post but now have issues with no.2

2) A uniform wooden beam 6 m long weighing 80 kg is lying horizontally on the floor. A builder raises one end until the beam is inclined at 300 to the horizontal and holds it in this position until his mate arrives to lift the other end. What is the magnitude of the force he exerts?

This was my attempt:

T=Fdsinθ, 80*9.8(784)*3m*sin30 = 1176

Then F = Td/sinθ, where 1176/sin30*6 = 392N but this was wrong

19. Nov 30, 2013

### voko

Why is that sin 30?