a uniform pole 20ft long ang weighing 50lbs is used to carry a load of 180lbs.Two men,one supporting twice as much as the other support the pole to keep it in a horizontal position.How far from the center of the pole should the load be hung.?

sum of forces:

50+180=f1+F2 where f2 and f1 are the men pushing upward.


let f1=2F2 (given) f3 is the lighter load man.

50+180=3F2
f2=230/3
f1=460/3

sum moments (about the center of pole).
10F1-10F2+x180=0
you know f1, f2, solve for x

To determine the distance from the center of the pole where the load should be hung, we need to consider the torques on the pole created by the load and the two men supporting it.

A torque is the rotational equivalent of a force. It is calculated by multiplying the force applied to an object by the perpendicular distance from the point of rotation (in this case, the center of the pole).

Let's assume that the two men are supporting the pole at distances d1 and d2 from the center of the pole. Since one man is supporting twice as much weight as the other, let's say that the force applied by the first man is F and the force applied by the second man is 2F.

The total torque acting on the pole must be zero for it to remain in a horizontal position, i.e., the torques created by the load and the two men should cancel each other out.

Mathematically, we can represent this as:

Torque due to the load = Torque due to the first man + Torque due to the second man

Now, let's calculate the torques for each component:

Torque due to the load:
Since the load is hung at distance d from the center of the pole and weighs 180 lbs, the torque due to the load is given by:

Torque_Load = Load * distance = 180 lbs * d

Torque due to the first man:
Since the first man is supporting twice as much weight as the second man, his force is 2F. Let's say he is supporting the pole at a distance d1 from the center. The torque due to the first man is given by:

Torque_FirstMan = 2F * d1

Torque due to the second man:
The second man is supporting the pole at a distance d2 from the center. Let's say his force is F. The torque due to the second man is given by:

Torque_SecondMan = F * d2

We know that the total torque acting on the pole is zero, so:

Torque_Load = Torque_FirstMan + Torque_SecondMan

Substituting the values we have:

180 * d = 2F * d1 + F * d2

Now, we know that the pole is in equilibrium when supported at the center, which means that the center of the pole should be in the middle of the two men:

d1 + d2 = 20 ft / 2 = 10 ft

We can rewrite this equation as:

d2 = 10 ft - d1

Substituting this into the previous equation, we get:

180 * d = 2F * d1 + F * (10 ft - d1)

Now, we need one more piece of information to solve the problem: the values of F and d1.