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marta [7]
3 years ago
15

An elevator car, with a mass of 450 kg is suspended by a single cable. At time = 0s, the elevator car is raised upward. The tens

ion on the cable is constant at 5000N during the first 3 seconds of operation. Determine the magnitude of the velocity of the elevator at time = 3 seconds.
Physics
1 answer:
Alexeev081 [22]3 years ago
5 0

Answer:

33.33 m/s

Explanation:

m = 450 kg. T = 5000 N, t = 3 seconds,

let the net acceleration is a.

T = m a

a = 5000 / 450 = 11.11 m/s^2

u = 0 , v = ?

Let v be the velocity after 3 seconds.

Use first equation of motion

v = u + a t

v = 0 + 11.11 x 3 = 33.33 m/s

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Answer:

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Explanation:

From the question we are told that

     The diameter of the shaft steel is  d =  50mm

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            P_B = \frac{\delta _d E (D^2 - d^2)}{2 * d* D^2}

Substituting values

                P_B = \frac{(0.005) (207 *10^{3} ) * (70^2 - 50^2))}{2 * (50) (70) ^2 }

                 P_B = 5.069 MPa

Now he axial force required is

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Where A is the area which is mathematically evaluated as

               \pi d l

So   P  =  \mu P_B \pi d l

Substituting values

      P =  0.2 * 5.069 * 3.142 * 50 * 100

       P =  15.93 k N

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