A lifting system can be modelled as two blocks of mass m1 and m2 hanging from two pulleys, connected by ropes to a third block of mass M = 100 kg, lying on top of a horizontal surface. There is gravity g = 9.80 m/s2.
a) Ignoring at first friction, find an expression for the acceleration of the blocks, as a function of the masses and gravity.
b) Now, we assume that the system starts at rest, and that there is static friction between the block M and the surface it is lying on. How much diffe- rence in mass can m1 and m2 have, for the system to not start moving, if the coefficient of static friction is μs = 1.00? Provide an expression as well as a numerical value.
c) Next, we assume that the system is moving with an initial speed vi = 2.00 m/s, when suddenly kinetic friction between M and the surface is triggered (the brakes are switched on). The coefficient of kinetic friction is μk = 0.700. How long does it take for the system to stop if m2 = m1 = 100 kg ? Provide an expression as well as a numerical value.
ΣF = ma
F* = F
The Attempt at a Solution
I think I did a correctly, but I have some problems solving b.
I put positive direction downwards to the left.
a) Firstly I separated the system into 3 systems:
1) am1 = m1g - T1
2) am2 = T2 - m2g
3) This one is the third block, lying on the top of the surface.
am3 = T1 - T2
then add together:
so that: a (m1 + m2 + m3) = m1g - T1 + T2 -m2g + T1 - T2
a = (m1 + m2)*g / (m1 + m2 + m3)
I know that μs = 1.00.
Fs = nμs = mgμs = 100kg * 9.80 * 1.00 = 980N
That means that m1g - m2g = 980N
Then divide by g:
m1 - m2 = 100kg
Is that the answer? That the difference in mass has to be less or equal to 100kg?