Phil takes the elevator down from the skyboxes. He makes some measurements while
ID: 2052642 • Letter: P
Question
Phil takes the elevator down from the skyboxes. He makes some measurements while riding down the elevator. He stands on a scale and notes various readings. He has mass of 78 kg. The mass of the elevator and the other occupants is 820 kg.A) As the elevator starts down, the scale Phil is standing on reads 447 N. Determine the magnitude of the acceleration of the elevator. (m/s2)
B) Determine the tension in the cable supporting the elevator as the elevator starts down. (N)
C) As the elevator nears the bottom of the skyboxes and comes to a stop, Phil notes that the scale he is standing on reads 996 N. Determine the magnitude of the acceleration of the elevator as it comes to a stop. (m/s2)
D) Determine the tension in the cable supporting the elevator as the elevator comes to a stop. (N)
E) He notes that the initial acceleration as the elevator starts down lasts for 1.0 seconds and that the elevator goes at a constant speed for 8.9 seconds before it begins to slow down at the bottom. Phil does a quick calc, and determines the distance the elevator traveled. What number did Phil calculate, assuming Phil worked the problem correctly? (m)
Explanation / Answer
a) when the elevator starts down the actual force on phil is F = m*(g-a) = 447N
=> a= 4.07 m/s^2.
b)since we know m*g - T = m*a.
where T= tension in the cable.
where M =mass of the elevator including phi = 820+78 = 898kg
=> T =M*(g-a) = 898*(5.73) = 5146.23 N.
c) while coming to stop the elevator is slowed down and a deceleration is produced
d.
now the deceleration is given by m*(g+d) = 996N
=> d= 2.95 m/s^2.
d) now the tension in the cable is given by
T - M*g = M*d
=> T = M*(g+a).
=> T = 898*(12.77) = 11.466 *10^3 N.
e) the time for which the lift accelerates t1 = 1 sec.
=> the distance traveled by the lift is d1 = 1/2*a*t1^2 = 1/2*4.07*1 = 2.035 mts.
now the distance covered during zero acceleration period
d2 = v*t2 = 4.07*8.9 = 36.223 mts.
now the distance covered during deceleration d3 = v*t3- 1/2*d*t3^2.
where t3 = v/d = 4.07/2.95 = 1.37 secs.
now d3 = 5.615 - 1/2*2.95*(1.37^2) = 2.846 mts.
the total distance travelled by the elevator before coming to rest is = d1+d2+d3 = 41.14 m.