A classical mechanics problem by Pankaj Joshi

Classical Mechanics Level 3

In the figure, the block of mass M (=10 kg) is at rest on the floor. The minimum acceleration with which a boy of mass m (=2 Kg) should climb along the rope of negligible mass so as to lift the block from the floor is ( g = 10 m / s 2 g= 10m/s^2 ):

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Pankaj Joshi by Pankaj Joshi , 23, India

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3 solutions

R.L. Bhat
Jul 6, 2014

when the boy hangs on the rope, he exerts a force of 20 N. To lift the block 100 N force is required. The additional force of 80 N if provided by accelerating with 40 m/s^2.

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IF somebody can explain please because Im totally messed up right now,what I thought is 50 ms^-2 the right answer,Please explain

Mahmoud Fathy - 6 years, 11 months ago

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If the boy hangs on the rope, he will exert 20 N force which is not sufficient to rise the 10 kg mass.It requires 100 N force. The additional 80 N force is obtained by accelerating upwards with 40 m/s^2.

R.l. Bhat - 6 years, 11 months ago

Same is the case here...

Pramita Kastha - 5 years, 10 months ago

Really nice solution sir.

Department 8 - 5 years, 10 months ago

I don't think (from where the boy stands) that he should climb up along the rope , but instead he should perform an acceleration of (40 m/s^2) by pulling the rope downwards while he is hanging .. Please anyone correct me if I'm wrong ??

Mohamed Ahmed Abd El-Fattah - 5 years, 10 months ago
Achille 'Gilles'
Aug 14, 2015

Using F=mg, (Newton's 2nd law)

Left side; 10 kg x 10 m/s² = 100 N

Right side, mass of the climber x g =

2 kg x 10 m/s² = 20 N

So if I need 80 N more with a climber of a mass of 2 kg,

80 N / 2 kg = 40 m/s²

Of course, a little more acceleration will be needed to lift the block as 40 m/s² is what needed to keep the system in balance.

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T T tension G G gravity F F total

On the Left side: F = T − G = m ⋅ a = 0 N F=T-G=m\cdot a=0N Then T = G = 10 ⋅ 10 = 100 N T=G=10\cdot 10=100N On the Right side: F = T − G = m ⋅ a F=T-G=m\cdot a Then a = T − G m = 100 − 2 ⋅ 10 2 = 40 m / s 2 a=\frac{T-G}{m}=\frac{100-2\cdot 10}{2}=40m/s^2

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