A block of mass M is pulled along a horizontal frictionless surface by a rope of mass m, as shown in Fig. 5-35. A horizontal force acts on one end of the rope.

(a) Show that the rope must sag, even if only by an imperceptible amount.
(b) Then, assuming that the sag is negligible, find the acceleration of rope and block.
(c) Find the force on the block from the rope.
(d) Find the tension in the rope at its midpoint.

Explanation

(a) Show that the rope must sag, even if only by an imperceptible amount.
Reasoning: A small segment of the rope has mass and is pulled down by the gravitational force of the Earth. Equilibrium is reached because neighboring portions of the rope pull up sufficiently on it. Since tension is a force along the rope, at least one of the neighboring portions must slope up away from the segment we are considering. Then, the tension has an upward component, which means the rope sags.
(b) Then, assuming that the sag is negligible, find the acceleration of rope and block.
The only force acting with a horizontal component is the applied force `vec(F)`. Treat the block and rope as a single object.
Calculation: We write Newton.s second law for it: `F=(M+m)a`, where a is the acceleration and the positive direction is taken to be to the right. The acceleration is given by `a=F//(M+m)`.
(c) Find the force on the block from the rope.
Calculation: The force of the rope `F_(r)` is the only force with a horizontal component acting on the block. Then Newton.s second law for the block gives
`F_(r)=Ma=(MF)/(M+m)`
where the expression found above for a has been used. (d) Find the tension in the rope at its midpoint.
Calculation: Treating the block and half the rope as a single object, with mass `M+1//2m`, where the horizontal force on it is the tension `T_(m)` at the midpoint of the rope, we use Newton.s second law:
`T_(m)=(M+(1)/(2)m)a`
`=((M+m//2)F)/((M+m))`
`=((2M+m)F)/(2(M+m))`.