two men are carrying a 12m uniform ladder on their shoulders. Man A is 1.0m from one end, and man B is 3.0m from the other end. If the beam has a mass of 200kg, determine the load supported by each man.
two motorcycles of equal mass collide at a 90 deg. intersection. if the momentum of motorcycle A is 450 kg km/h west and the momentum of motorcycle B is 725 kg km/h south, what is the magnitude of the resulting momentum of the final mass?
To determine the load supported by each man, we need to consider the distribution of weight along the ladder. Since the ladder is uniform, we can assume the weight is distributed evenly.
First, let's calculate the center of mass of the ladder. Since the ladder is 12m long and assumed to be uniform, the center of mass is located in the middle of the ladder, which is 6m from each end.
Now, let's consider the position of the two men. Man A is 1.0m from one end, which means he is 6 - 1 = 5.0m from the center of mass. Similarly, Man B is 3.0m from the other end, which means he is 6 + 3 = 9.0m from the center of mass.
To calculate the load supported by each man, we need to find the ratio of their distances from the center of mass.
The load supported by each man can be calculated using the following formula:
Load A = (distance from A to center of mass) / (total ladder length) * (total mass)
Load B = (distance from B to center of mass) / (total ladder length) * (total mass)
Plugging in the values, we get:
Load A = (5.0m / 12m) * (200kg) = 83.33kg (rounded to two decimal places)
Load B = (9.0m / 12m) * (200kg) = 150kg
Therefore, Man A supports a load of approximately 83.33kg, and Man B supports a load of 150kg.