A cable in which there is 12500 N of tension force supports an elevator. What is the magnitude and direction of the acceleration of the elevator if its total mass is 1175 kg?

a = 10.638m/s^2 in a vertical direction, up 90 degrees from the x axis

Explanation:

Givens:

F_t = 12500N

w = 1175kg

g = -9.81m/s^2

F = m(a)

F_w = m(g)

So taking the weight of the elevator and multiplying it with gravity we get:

F_w = 1175(-9.81)

F_w = -11526.75N

So since the weight of the elevator and the acceleration of gravity would result in 11526.75N of force down the cable would have the opposite force pulling up 11526.75N but it doesn't it has more force then that so the elevator is producing enough force to defeat gravity and move upwards. So now we take our total force and subtract the force of the weight on the cable.

12500N-11526N = 973.25N

So the elevator after defeating gravity is still producing 973.25N of force upwards. So now we take the equation for force and rearrange it:

F = m(a)

a = F/m

a = 0.828m/s^2

But this is in addition to countering the gravity so add 9.81m/s^2 to this and you get

a = 10.638m/s^2 in a vertical direction, up 90 degrees from the x axis

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s

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