Physics Lecture 3

Consider the case of a set of boxes, all of them equal size, equal mass m, and connected by massless strings. If another massless string is attached to one of the end boxes and used to accelerate the boxes vertically upward with acceleration a, what is the tension in this string for a set of boxes totalling a length l (assume that the strings between boxes all have equal length)?

Solution

The first thing to note is that if the strings don't stretch, then the accelerations of all the boxes must be equal, hence a1 = a2 = a3. The tensions will not be equal since the forces acting on each box are different. To find out what the forces are, let's look at just three boxes for a start. If choose to look at the bottom box, then
T3 - mg = ma ==> T3 = m(g+a)

For the second (or middle) box, the equation of motion must take into account that the box is being pulled up by tension T2 but down by the tension in the rope attached to both the second and third boxes, so

T2 - T3 - mg = ma ==> T2 = T3 + m(g+a) ==> T2 = 2m(g+a)

The top box is pulled up by T1 and down by the tension in the string attaching the top and middle boxes, thus

T1 - T2 - mg = ma ==> T1 = 3m(g+a)

Now you get the pattern, don't you? Each box adds a tension m(g+a) to the string pulling the box above it. If we set the length of each box and string pulling it to L, then the number of boxes must be l/L. For example, if the length of a box and string pulling is L = 10 cm and we had a total length, l, of 50 cm, then the number of boxes would 5. The total mass of all the boxes would be ml/L, so the tension in the topmost box is

T = ml/L * (g+a)
Since we often have problems involving a calculation of mass like this, we have a standard name for the quantity m/L or mass per unit length: we call it the linear mass density. The concept of density is used to solve a great number of problems.

Suppose we extend the argument to the case of a rope with mass connecting two boxes held in the familiar arrangement shown below. What is the acceleration of the two boxes when they are released?

The boxes have masses m1 and m2 as shown. The rope has a total mass M and length l. The distance y represents the amount of rope hanging over the edge of the table top. Initially the boxes are held at rest and y = d where d is the amount of rope initially hanging over the edge. The forces are as shown below

Using these forces, we can write down the equations of motion and solve to find the acceleration. There is a Maple file below that does just this.

Solution to massive rope problem.