The elevator and passengers have a mass of 5500 kg. After a stop at the thirteenth floor, the elevator accelerates upward uniformly at 1.0 m/s^2 until it is rising at 2.0 m/s. What is the magnitude of the net force acting on the elevator as it accelerates?
Draw a free body diagram about the elevator and sum vertical forces
So T - m*g = m*a so T (net force) = m*(g + a) = 5500*(9.8 + 1.0) = 59400N
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April 7th, 2010 at 10:49 am
The average acceleration the elevator experiences while accelerating is 1.5 m/s², because it is uniform.
F = ma = 5500 kg * 1.5 m/s² = 8250 N
Direction of the force is UP.
References :
April 7th, 2010 at 10:54 am
Draw a free body diagram about the elevator and sum vertical forces
So T - m*g = m*a so T (net force) = m*(g + a) = 5500*(9.8 + 1.0) = 59400N
References :