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## Homework Statement

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.

## Homework Equations

ΣF = ma

F* = F

(Newton´s laws)

## 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) a

_{m1}= m

_{1}g - T

_{1}

2) a

_{m2}= T

_{2}- m

_{2}g

3) This one is the third block, lying on the top of the surface.

a

_{m3}= T

_{1}- T

_{2}

then add together:

so that: a (m

_{1}+ m

_{2}+ m

_{3}) = m

_{1}g - T

_{1}+ T

_{2}-m

_{2}g + T

_{1}- T

_{2}

a = (m

_{1}+ m

_{2})*g / (m

_{1}+ m

_{2}+ m

_{3})

b)

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?

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