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serg [7]
3 years ago
13

In a pulley system, two blocks are connected by a rope as shown below. The coefficient of static friction between block A and th

e surface is 0.8. Block A has a mass of 20kg. What is the minimum mass that block B can have so that block A moves?
Physics
1 answer:
umka21 [38]3 years ago
5 0

Hi there!

We can begin by doing a summation of forces for each block.

Block A:

This block has the force of tension (in direction of acceleration +) and static friction (opposite direction -) acting on it. Thus:

\Sigma F_A = T - F_s\\\\m_Aa = T - \mu m_Ag

Block B:

This block has the force of tension (opposite of acc. -) and gravity (in direction of acc +), working on it.

\Sigma F_B = m_Bg - T\\\\ m_Ba = m_Bg - T

Add both of the expressions and solve for the maximum mass of Block B.

\Sigma F = m_Bg - T + T - \mu m_Ag\\\\a(m_A + m_B) = m_Bg - \mu m_Ag

To find the minimum value, we can set a = 0, so:

0 = m_Bg - \mu m_Ag\\\\\mu m_Ag = m_B g\\\\0.8(20)(9.8) = m_B (9.8)\\\\m_B = \frac{0.8(20)(9.8)}{9.8)} = \boxed{16 kg}

The block must weigh <u>> 16 kg</u> for block A to move.

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(1 point) A rectangular tank that is 3 feet long, 9 feet wide and 12 feet deep is filled with a heavy liquid that weighs 110 pou
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Answer:

Explanation:

Work in pumping water from the tank is given as

W = ∫ y dF. From a to b

Where dF is the differential weight of the thin layer of liquid in the tank, y is the height of the differential layer

a is the lower limit of the height

b is the upper limit of the height.

We know that, .

F = ρVg

Where F is the weight

ρ is the density of water

V is the volume of water in tank

g is the acceleration due to gravity

Then,

dF = ρg ( Ady)

We know that the density and the acceleration due to gravity is constant, also the base area of the tank is constant, only the height that changes.

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Area = L×B = 3 × 9 = 27ft²

dF = ρg ( Ady)

dF = 1684.8dy

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W = ∫ y dF. From a to b

W = ∫ 1684.8y dy From 0 to 12

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3 years ago
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