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Question

ofradrad/swithartsiding. (3pe(1point, of the tire --(1 point). 16. A rolls at Find the car's velocity Total linear yelocity of thetop part and total linear velocity the tire of the bottom part of 17. Push-ups problem. (2 points) An athlete of mss 90 kg and height 2 m is doing a series of pushups as figure below. His center kg and height 2 m is doing a series of pushups as shown in the his palms are 30.0 cm (-0.3 m) from the top of his head. of mass is 1.2 m from the bottom of his feet, and the centers of 20m 30.0 cm Find the force that the floor exerts on each of his feet, assuming that both feet exert the same force. Find the force that the floor exerts on each of his hand, both palms exert the same force

Explanation / Answer

The man’s hands and feet are in contact with the floor. So, the weight of the person is supported at these two positions.
Let F1 be the upward force which the floor exerts on his feet; and F2 be the upward force which the floor exerts on his hands.

Weight = 90 * 9.8 = 882 N
F1 + F2 = 882

As the man pushes downward, his body rotates counter clockwise. As he relaxes his muscles, he rotates clockwise. The pivot point is the point where his feet touch the floor. The upward force, which the floor exerts on his hands, causes the counter clockwise torque. His weight causes the clockwise torque.
The distance from his feet to his center of mass is 1.2 m. The distance from his feet to the centres of his palms is 2 – 0.30 = 1.7 m

Counter clockwise torque = 882 * 1.2 = 1058.4 n.m
Clockwise torque = F2 * 1.7
F2 * 1.7 = 1058.4
F2 = 622.6 N

F1 = 882 – 622.6 = 259.4 N

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