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Kazeer [188]
4 years ago
8

The helicopter in the drawing is moving horizontally to the right at a constant velocity. The weight of the helicopter is W=5240

0 N. The lift force L generated by the rotating blade makes an angle of 21.0° with respect to the vertical. (a) What is the magnitude of the lift force? (b) Determine the magnitude of the air resistance R that opposes the motion.

Physics
1 answer:
pantera1 [17]4 years ago
3 0

Explanation:

It is given that,

The weight of the helicopter, W = 52400 N

The lift force L generated by the rotating blade makes an angle of 21.0° with respect to the vertical. The attached figure shows the free body diagram of the helicopter.

(a) Let L is the magnitude of the lift force.

The net force acting in the vertical direction,

L\ cos\theta-W=0

L=\dfrac{W}{cos\ \theta}

L=\dfrac{52400}{cos(21)}

L = 56127.99 N

L\ cos[tex]L=5.61\times 10^4\ N

(b) The net force acting in the x direction is :

L\ sin\theta-R=0

R=L\ sin\theta

R=5.61\times 10^4\times \ sin(21)

R=2.01\times 10^4\ N

Hence, this is the required solution.

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Two blocks are suspended from opposite ends of a light rope that passes over a light, frictionless pulley. One block has mass m1
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Answer:

a)  m₁ = 1.41 kg , b)  m₂ = 2.65 kg

Explanation:

For this exercise we will use Newton's second law

Block 1

             T - W₁ = m₁ a

Block 2

           W₂ - T = m₂ a

We have selected the positive  block 1 rising and block two lowering, as the pulley has no friction does not affect the movement

Let's use kinematics to look for acceleration

         y = v₀ t + ½ a t²

As part of the rest the initial speed is zero

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Let's replace in the equation of block 1

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b) we substitute in the equation of block 2

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     m₂ = T / (g-a)

     m₂ = 18.0 / (9.8 -3)

    m₂ = 2.65 kg

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