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Alexeev081 [22]
4 years ago
10

A house painter uses the chair-and-pulley arrangement of (Figure 1) to lift himself up the side of a house. The painter's mass i

s 55 kg and the chair's mass is 10 kg .
If the painter is at rest, what is the tension in the rope?
Physics
1 answer:
Annette [7]4 years ago
6 0

Answer:

Explanation:

Draw a free body diagram for the painter.  There are three forces acting on him.  The tension in the rope pulling him up, the normal force of the chair pushing him up, and gravity pulling him down.

Apply Newton's second law:

∑F = ma

T + N - W = ma

Since the painter is at rest, a = 0.

T + N - W = 0

T = W - N

T = Mg - N

Now draw a free body diagram of the chair.  The chair has three forces acting on it also.  The tension in the rope pulling it up, the normal force pushing down, and gravity pulling down.

Newton's second law for the chair:

∑F = ma

T - W - N = ma

Since the chair is also not accelerating, a = 0:

T - W - N = 0

T = W + N

T = mg + N

Now we have two equations and two variables.  If we add the equations together, we can eliminate N:

2T = Mg + mg

2T = (M + m) g

T = (M + m) g / 2

Given that M = 55 kg, m = 10 kg, and g = 9.8 m/s²:

T = (55 + 10) (9.8) / 2

T = 318.5 N

If we round to two sig-figs, the tension in the rope is 320 N.

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Answer:

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Explanation:

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Let m be the mass must be on the other end if the plank remains balanced.

In balance condition

20\times 1=10\times (1.5-1)+m\times (1.5+0.5)

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2m=20-5=15

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what is refractive index?

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Answer:

Therefore the terminal velocity = 1.45 m/s

Explanation:

Terminal velocity: Terminal velocity is the highest velocity of an object when it falls from rest trough a media.

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V_t= terminal velocity

w = weight of the object = mg

c_d = drag coefficient=0.80

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Answer:

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