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ziro4ka [17]
3 years ago
7

A 975-kg elevator accelerates upward at 0.754 m/s2, pulled by a cable of negligible mass. Find the tension force in the cable.

Physics
1 answer:
Zina [86]3 years ago
7 0

To solve this problem we will apply the concepts of equilibrium and Newton's second law.

According to the description given, it is under constant ascending acceleration, and the balance of the forces corresponding to the tension of the rope and the weight of the elevator must be equal to said acceleration. So

\sum F = ma

T-mg = ma

Here,

T = Tension

m = Mass

g = Gravitational Acceleration

a = Acceleration (upward)

Rearranging to find T,

T = m(g+a)

T = (975)(9.8+0.754)

T= 10290.15N

Therefore the tension force in the cable is 10290.15N

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Complete Question

An athlete at the gym holds a 3.0 kg steel ball in his hand. His arm is 70 cm long and has a mass of 4.0 kg. Assume, a bit unrealistically, that the athlete's arm is uniform.

What is the magnitude of the torque about his shoulder if he holds his arm straight out to his side, parallel to the floor? Include the torque due to the steel ball, as well as the torque due to the arm's weight.

Answer:

The torque is  \tau = 34.3 \  N\cdot m

Explanation:

From the question we are told that

   The mass of the steel ball is  m  =  3.0 \  kg

    The length of arm is  l =  70 \ cm  = 0.7 \  m

    The mass of the arm is m_a  = 4.0 \  kg

Given that the arm of the athlete is uniform them the distance from the shoulder to the center of gravity of the arm is mathematically represented as

       r = \frac{l}{2}

=>    r = \frac{ 0.7}{2}  

=>    r = 0.35 \ m  

Generally the magnitude of torque about the athlete shoulder is mathematically represented as

      \tau =  m_a * g * r  + m * g *  L

=>    \tau =  4 * 9.8 * 0.35 + 3 * 9.8 *  0.70

=>    \tau = 34.3 \  N\cdot m

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A 55 kg person falling with a velocity of 0.6
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Answer:

What are we supposed to find, if it is kinetic energy then this is the solution.

K.E=1/2mv^2

K.E= kinetic energy

M=mass

V=velocity

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Explanation:

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Explanation:

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Answer:

Force on front axle = 6392.85 N

Force on rear axle = 8616.45 N

Explanation:

As we know that the weight of the car is balanced by the normal force on the front wheel and rear wheels

Now we know that

F_1 + F_2 = W

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now we know that distance between the axis is 2.70 m and centre of mass is 1.15 m behind front axle

so we can write torque balance about its center of mass

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