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Paladinen [302]
2 years ago
6

) A friend of yours complains that he often has lower back pain. One day while he picks up a package, you notice he bends at his

hips rather than keeping his back straight and bending his knees. Assume he pivots at the hips so that his spine is horizontal. His back muscles then exert a force at 48 cm away from the hips at 13.5 degrees above the horizontal (directed slightly up and to the left in the picture shown). The weight of his upper body is 560 N, and his center of mass is 38 cm from the hips. Draw a force diagram of the spine. What total force is exerted by the back muscle when he picks up a 10 kg package at 82 cm from the hips? Extra credit: Find the force exerted by the hips on the spine/pelvis
Physics
1 answer:
max2010maxim [7]2 years ago
3 0

Answer:

Explanation:

Please see the figure attached for free body diagram .

Taking torque about the hip joint ,

560 x 38 + 100 x 82 = F sin13.5 x 48

21280 + 8200 = 11.2 F

F = 2632.14 N .

Force exerted by the hip on the spine = 2632.14 N .

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The rocket is fired vertically and tracked by the radar station shown. When θ reaches 66°, other corresponding measurements give
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Answer:

velocity = 1527.52 ft/s

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Explanation:

We are given;

Radius of rotation; r = 32,700 ft

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Also, angle θ reaches 66°

So, velocity of the rocket for the given position will be;

v = rθ˙˙/cos θ

so, v = 32700 × 0.019/ cos 66

v = 1527.52 ft/s

Acceleration is given by the formula ;

a = a_r/sinθ

For the given position,

a_r = r¨ - r(θ˙˙)²

Thus,

a = (r¨ - r(θ˙˙)²)/sinθ

Plugging in the relevant values, we obtain;

a = (85 - 32700(0.019)²)/sin66

a = (85 - 11.8047)/0.9135

a = 80.13 ft/s²

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3 years ago
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Explanation:

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Answer:

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Therefore, the energy stored in the inductor is 10t^2e^-10t Joules

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