Q:1To strengthen his arm and chest muscles, an 81.0 kgathlete 2.00 m tall is doi
ID: 1490163 • Letter: Q
Question
Q:1To strengthen his arm and chest muscles, an 81.0 kgathlete 2.00 m tall is doing a series of push-ups as shown in the figure below (Figure 1) . His center of mass is 1.10 m from the bottom of his feet, and the centers of his palms are 30.0 cm from the top of his head. Find the force that the floor exerts on each of his feet and on each
a.Find the force that the floor exerts on each of his feet, assuming that both feet exert the same force.
b.Find the force that the floor exerts on each of his hand, both palms exert the same force.
Q:2 60.0-cm, uniform, 49.0-N shelf is supported horizontally by two vertical wires attached to the sloping ceiling (the figure (Figure 1) ). A very small 19.0-N tool is placed on the shelf midway between the points where the wires are attached to it.
a.Find the tension in left wire.
b.Find the tension in right wire
Explanation / Answer
Q:1
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 = 81 * 9.8 = 793.8
F1 + F2 = 793.8
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 1.10 m. The distance from his feet to the centers of his palms is 2 – 0.30 = 1.7 m
Counter clockwise torque = 793.8 * 1.10 = 873.18
Clockwise torque = F2 * 1.7
F2 * 1.7 = 873.18
F2 = 873.18 ÷ 1.7
= 513.63
F1 = 793.8 – (873.18 ÷ 1.7)
=280.16
Q:2
recognize problem is one of Static Equilibrium of Forces acting on uniform shelf.
Convert physical diagram into a Force sketch on the Free-Body of the uniform shelf.
Make sure ALL forces appear on UR F-B diagram of this shelf:
2 forces of Tension - one in each supporting wire and 2 weight forces = 4 forces in all
T1 = left wire's tension force
T2 = right wire's tension force
small tool's weight = 19 N {acts cm from left hand end of shelf}
shelf weight = 49 N (acts 40 cm from left hand end of shelf}
F = 0 {vertical direction}
T1 + T2 = 19 + 49 = 68 N
T1 = 68 - T2
Now apply Static Equilibrium to all Torques (moments) acting on shelf,
take LEFT-END (or UR choice) of shelf location as PIVOT point for torques:
let CW torque be positive
torques = 0 = (19)(0.3) +(49)(0.4) - (0.6)T2
0.6T2 = 5.7 + 19.6 = 25.3
T2 = 25.3/0.6 = 42.16 N ANS
T1 = 68 - 42.16 = 25.84 N ANS
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