The diagram shows an inclined elevator platform (the darker triangle the person
ID: 1659579 • Letter: T
Question
The diagram shows an inclined elevator platform (the darker triangle the person is standing on), attached by a cable that passes over a pulley and is tied to a block. The platform can move along the surface of the ramp, which is fixed in place. The angle between the surface of the ramp and the horizontal is 30º. The mass of the platform is M= 270 kg. The person on the platform has a mass of 55.0 kg. Use g = 10 m/s2, and assume the pulley has negligible mass and rotates without friction between the pulley and its axle.
(a) First, assume there is no friction between the platform and the ramp. Calculate the mass the hanging block should have to keep the system from moving, if the system starts from rest.
____ kg
It turns out that there is some friction between the platform and the ramp. The coefficient of static friction is 0.600 and the coefficient of kinetic friction is 0.400.
(b) If the system starts from rest now, what is the largest value the mass of the hanging block can have without the system moving?
___kg
(c) With the mass of the block at the value from part (b), someone comes along and gives the platform a brief push, so that it starts moving up the ramp. With that person no longer pushing the platform, what will the magnitude of the platform's acceleration be?
___ m/s/s
(d) Calculate the magnitude of the tension in the cable, for the situation described in part (c).
___ N
Explanation / Answer
a) the mass of block should be such that its weight balances the weight of man+platform acting along the incline
mg=(270+55)*g*sin30
Mass of hanging block = m=162.5 kg
b)The mass the hanging block can have = m
mg=(270+55)*g*sin30 + 0.6*(270+55)*g*cos30
m=331.4 kg
c)once the initial push is given, kinetic friction will be mobilised.
From free body diagram of hanging block,
331.4*10 - T =331.4a....(i)
From free body diagram of platform and man,
T-(325*10*sin30)-(0.4*325*10*cos30) = 325a
Or,T-1625-1125.8=325a
Or, T-2750.8=325a....(ii)
Solving above two equations,we get
a=0.86m/s²
Magnitude of acceleratiom=0.86m/s²
d)Tension on cable=T=3029.7 N