Question

In order to raise a block of mass $ 100{\mkern 1mu} kg $ a man of mass $ 60{\mkern 1mu} kg $ fastens a rope to it and passes the rope over a smooth pulley. He climbs the rope with an acceleration $ \dfrac{{5g}}{4} $ relative to the rope. The tension in the rope is ( $ g = 10{\mkern 1mu} m/{s^2} $ ).

Answer 1

Hint: We will first use Newton’s second law of motion twice on the two bodies separately. The two bodies in consideration here are the man and the block. Once the two equations are found, we will solve them to get the required tension since there are only two unknowns here, i.e. the acceleration of the rope and the Tension in the rope.

Complete Step by step solution
The only forces acting on the man are those of the tension and the gravitational attraction force of the planet. The tension of the rope is being generated by his pull with which he is climbing the rope which in turn is raising the block. The Tension in the rope is acting upwards for both the man and the mass and this force helps the man to climb the rope against the gravity.
Let us assume the tension in the rope to be $ T $ . Let the acceleration of the rope and in-turn the block be $ a $ such that the block moves upwards.
Thus equating the forces on the block by applying Newton’s second law of motion, we get $ T - Mg = Ma $ ,
where $ M = 100{\mkern 1mu} kg $ is the mass of the block.
 $ \Rightarrow T = M(g + a) $
The acceleration of the man is given to be $ \dfrac{{5g}}{4} $ relative to the rope. Therefore its actual acceleration will be the relative acceleration less the acceleration of the rope since the rope is moving in the opposite direction.
Now equating the forces on the man by applying Newton’s second law of motion, we get $ T - mg = m(\dfrac{{5g}}{4} - a) $ .
 $ \Rightarrow T = m(\dfrac{{5g}}{4} - a + g) $
Equating the tension from the two equations, we get $ M(g + a) = m(\dfrac{{9g}}{4} - a) $
 $ \Rightarrow 100{\mkern 1mu} kg{\mkern 1mu} (g + a) = 60{\mkern 1mu} kg{\mkern 1mu} (\dfrac{{9g}}{4} - a) $
 $ \Rightarrow 5(g + a) = 3(\dfrac{{9g}}{4} - a) $
On further simplifying, we get
 $ \Rightarrow 5g + 5a = \dfrac{{27g}}{4} - 3a $
 $ \Rightarrow 8a = \dfrac{{7g}}{4} $
Dividing by $ 8 $ on both sides we get,
 $ \Rightarrow a = \dfrac{{7g}}{{32}} $
Now substituting this value of acceleration in the first equation of tension in the rope, we will get the required answer.
 $ \Rightarrow T = M(g + a) = M(g + \dfrac{{7g}}{{32}}) $
 $ \Rightarrow T = 100{\mkern 1mu} kg{\mkern 1mu} (\dfrac{{39g}}{{32}}) $
Now substituting the value of the acceleration due to gravity ( $ g = 10{\mkern 1mu} m/{s^2} $ ) in the equation, we get
 $ \Rightarrow T = 100{\mkern 1mu} kg{\mkern 1mu} (\dfrac{{39 \times 10{\mkern 1mu} m/{s^2}}}{{32}}) $
 $ \Rightarrow T = 1218.75{\mkern 1mu} N $
Therefore the tension in the rope is $ 1218.75{\mkern 1mu} N $ .

Note
The tension in the rope is assumed to be uniform. This is because the rope is massless as otherwise, the tension would also have to account for the mass of the rope. Here the rope lifts the block and so it moves in the downward direction on the side of the climbing man. Thus while calculating the absolute acceleration of the man, we will need to subtract the acceleration of the rope from $ \dfrac{{5g}}{4} $ .