Two objects of mass m1 = 12 kg and m2 = 18 kg are connected by a light cord and hung

over a frictionless pulley. Both cord and pulley have negligible mass. Find the tension in

the cord.

The pulleys are on opposite sides with m2 being on the right and m1 being on the left. I know for a fact that the heavier mass will move down and drag the other with it so you don't add them. The answer is 141 N. I keep getting roughly 60(58.8 precisely)

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• Anonymous
1 month ago

acceleration a = g*(m2-m1)/(m1+m2) = 9.806*(18-12)/(18+12) = 9.80/5 m/sec^2

tension t = m2*(g-a) = 18*4*9.80/5 = 141 N

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• More than one way to do this.

Fnet = (18-12)*9.8 = 58.8N

a = Fnet/m = 58.8/30 = 1.96m/s²

BUT the masses are ACCELERATING.

On the falling side:

T = m(g-a).  If a = 0, T = mg and if a = g, T = 0

T = 18(9.8-1.96) = 141.12 N

OR on the rising side

T = 12(9.8+1.96) = 141.12 N

Same answer since the pulley is ideal.

OR

mg - T = ma for m = 18 when mg > T

18*9.8 - T = 18*1.96

OR

T - mg = ma for m = 12 when T > mg

T - 12*9.8 = 12*1.96

Hope this helps.

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• Okay so you are subtracting the weights. That might seem like something to do on first sight. Let's stop and decide if that makes sense when the blocks have equal mass.

It doesn't.

Of course, it also doesn't make sense if one block was very heavy and one very light.

When faced with a problem, always examine the "special and edge cases".

Draw the free body diagrams.

The net force on mass m2 is m2*g - T downwards.

m2*g - T = m2*a

The net force on m1 is upwards so the signs are reversed

T - m1*g = m1*a

So when you multiply 1st equation with -m1 and 2nd with m2

T = 2m1m2*g / (m1 + m2)

In other news, this is not hard to google AT ALL.

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• Draw diagram

18g - T = 18a

T - 12g = 12a

solve for T--->

T = 72g/5 = 141.1 N

• oldschool
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