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irina1246 [14]
3 years ago
7

A 12,000-N car is raised using a hydraulic lift, which consists of a U-tube with arms of unequal areas, filled with oil and capp

ed at both ends with tight-fitting pistons. The wider arm of the U-tube has a radius of 18.0 cm and the narrower arm has a radius of 5.00 cm. The car rests on the piston on the wider arm of the U-tube. The pistons are initially at the same level. What is the initial force that must be applied to the smaller piston in order to start lifting the car
Physics
1 answer:
Bond [772]3 years ago
7 0

Answer:

= 925.92 N

≅ 926N

Explanation:

Pressure due to car = pressure due to applied force  

12000/18^2 = Force / 5^2

force = 12000 * 25/ 324

= 925.92 N

For equilibrium

Pressure1 = Pressure2

A1F1 = A2F2

12000*pi*(5^2) = F2 ( pi)*(18^2)

so, F2 = Applied force to lift car = 925.92 N

Pascal's principle

Pressure1 = Pressure2

F1/A1 = F2/A2 (F=force and A=area)

A1 =Pi*(0.05)²

A2 =Pi(0.18)²

F2=12000

F1 = 12000*(0.05)² / (0.18)² = 926N

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3 0
4 years ago
2) A traffic light of weight 100 N is supported by two ropes as shown. Let T1 and T2 are the tensions.
GarryVolchara [31]

Hi there!

a.

We know that:

\Sigma F_y = 0 \\\\\Sigma F_x = 0

Begin by determining the forces in the vertical direction:

W = weight of traffic light

T₁sinθ = vertical component of T₁

T₂sinθ = vertical component of T₂

b.

The ropes provide a horizontal force:

T₁cosθ = Horizontal component of T1

T₂cosθ = Horizontal component of T2

Thus:

0 = T₁cosθ  - T₂cosθ

T₁cosθ = T₂cosθ

T₁ = T₂

c.

Since the angles for both ropes are the same, we can say that:

T₁ = T₂

Sum the forces:

ΣFy = T₁sinθ + T₁sinθ - W = 0

2T₁sinθ = W

d.

Now, we can begin by solving for the tensions:

2T₁sinθ = W

T_1 = T_2 =  \frac{W}{2sin\theta} = \frac{100}{2sin(37)} = \boxed{83.08 N}

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3 years ago
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