An elevator cab and its load have a combined mass of 1200 kg. Find the tension i

Question

An elevator cab and its load have a combined mass of 1200 kg. Find the tension in the supporting cable when the cab, originally moving downward at 10 m/.s, is brought to rest with constant acceleration in a distance of 35 m. 25300 N 21200 N 18400 N 13500 N 8300 N A free-body diagram is shown for the following situation: a force P pulls on a crate that is sitting on a rough surface I he force P is directed at an angle theta above the horizontal direction F_N represents the normal force on the crate. W = mg represents the weight of the crate, and represents the frictional force Which one of the following actions would result in an increase in the normal force? Decrease of the angle theta Increase of the magnitude of P Decrease of the coefficient of friction Decrease of the magnitude of W Neither of the above

Explanation / Answer

m = 1200 kg, u = 10 m/s , v = 0, s = 35 m

From kinematic equaitons

v^2 - u^2 = 2as

0 -10^2 = -2*a*35

a = 1.4286 m/s^2

Net force along verticla direction

T -mg = ma

T - 1200*9.81= 1200*1.4286

T = 13500 N

correct option is (D)

4) P is resolve into tow components

Px = Pcso(theta)

Py = Psin(theta)

Net force along vertical direction

FN+Psin(theta) = W

Normal force FN = W - Psin(theta)

COrrect option is (A)

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