Need help with these questions A.S.A.P

[sens_freak_18]

I really need some help solving these two questions. Any help is appreciated!

1) A tapered(non-uniform) power pole has a mass of 350 kg and a length of 12.2m. A cable attached to the smaller end has a tension of 1110N just as the end is about to ris from the ground. Where would the cable have to be attached in order for the pole to rise horizontally?

2)If a power-lifter was to lift 263.5kg mass over his head using a 1.3m long barwith weights on the end, (assuming the centre of gravity for the bar and weights is at the center) what is the force each hand exerts if they are both 0.27m from each end?

I'm not asking for the answers to be given to me, I just need some help setting up the questions, and also i can't figure out where I should put the pivot points. Thanks again for any help!

 

[Diane_]

General rule: The location of the pivot point is irrelevant to the final answer. in other words, you can put it anywhere you want and you'll get the same thing. This having been said, it's generally a good idea to put the pivot point at the location of an unknown force if you have more than one. If you only have one unknown, then put it at the location of a known one. The force at the location of the pivot cancels out of the torque equations.

 

[wais]

1) To solve this question, we need to use the concept of torque. Torque is the force that causes rotation and is calculated by multiplying the force by the distance from the pivot point. In this case, the pivot point is the point where the pole is attached to the ground.

Let's label the larger end of the pole as point A and the smaller end as point B. The cable is attached to point B with a tension of 1110N. We also know that the mass of the pole is 350 kg and its length is 12.2m.

To find the pivot point where the cable needs to be attached for the pole to rise horizontally, we can use the equation for torque: torque = force x distance.

Since the pole is in equilibrium (not moving), the net torque acting on it must be zero. This means that the torque caused by the cable tension must be equal and opposite to the torque caused by the weight of the pole.

We can calculate the torque caused by the cable tension by multiplying the tension force (1110N) by the distance from the pivot point (x) to point B (which is the length of the pole, 12.2m).

The torque caused by the weight of the pole can be calculated by multiplying the weight (350kg) by the distance from the pivot point (x) to the center of mass of the pole (which we can assume to be at the midpoint, 6.1m).

Setting these two torques equal to each other and solving for x, we get:

1110N x 12.2m = 350kg x 9.8m/s^2 x 6.1m

x = 2.71m

Therefore, the cable needs to be attached 2.71m from the larger end of the pole for it to rise horizontally.

2) To solve this question, we need to use the concept of center of mass. The center of mass is the point where the weight of an object is evenly distributed, and it is also the point where the object will balance.

In this case, the center of mass of the bar with weights is at the center of the bar (since it is symmetrical). The force each hand exerts can be calculated by using the equation for torque again.

We can label the two weights on the bar as point A and B, and the hands as points C and D. We know the mass of