Two men are carrying a 9.00-m telephone pole that has a mass of 115 kg. If the center of gravity of the pole is 3.00 m from the right end, and the men lift the poles at the ends, how much weight must each man support?
The man closest to the center of gravity must lift 2/3 of the pole's weight. The other guy lifts 1/3 of the weight. You can prove that with a moment balance.
The pole's total weight is 115 x 9.8 Newtons.
You take it from there
Use the formula for center of mass.
(∑ distance * mass) / total mass = center of mass
The masses are what each man is holding, the distance is where each man is from some origin. If you set the origin on one of the men, his distance * mass is 0. Since we know the center of mass is 6, we can solve for the mass. 2/3 for the man on the right and 1/3 for the man on the left.
To find out how much weight each man must support, we need to consider the distribution of weight along the pole. In this case, the weight is concentrated at the center of gravity, which is located 3.00 m from the right end of the pole.
First, let's calculate the weight of the telephone pole using the formula: weight = mass × gravity.
Given:
Mass of the pole (m) = 115 kg.
Acceleration due to gravity (g) = 9.8 m/s^2 (approximately).
Weight of the pole = 115 kg × 9.8 m/s^2 ≈ 1127 kg·m/s^2 or 1127 N.
Now, we need to find out how this weight is distributed between the two men. Since the men are lifting the pole at the ends, we can assume that the pole is balanced. This means that the weight is evenly distributed between the two ends.
Hence, each man must support half of the total weight.
Weight each man must support = (Weight of the pole) / 2 = 1127 N / 2 = 563.5 N.
Therefore, each man must support approximately 563.5 Newtons (N) of weight.