0.17 Phy1150: force and motion -- units of force  (Page 8/9)

P = Q = R = 49.05 newtons

First dynamic scenario

Now assume that someone cuts the cord that attaches Mass A to Mass B. What happens to the tension in each cord? Which, if any of the masses move, and ifthey move, what is their acceleration?

We probably don't need to do any calculations to answer these questions. Life experience tells us that the tension in each cord immediately goes tozero when the cord holding up the weight is cut.

Movement

Mass B and Mass C remain in equilibrium with no horizontal forces acting on them and their weights being supported by the table. They don't move.

Mass A immediately begins a free fall toward the floor with an acceleration that is equal to the acceleration of gravity at 9.81 m/s^2. A short segment ofcord trails out behind Mass A like an unopened parachute.

Second dynamic scenario

Now assume that we start over with the original static scenario and someone cuts the cord that attaches Mass A to Mass B. Whathappens to the tension in each cord? Which, if any of the masses move, and if they move, what is their acceleration?

This situation is a little more complicated and will probably require some calculations to sort out.

It seem obvious that the tensions labeled Q and R immediately go to zero, but the tension labeled P does not go to zero.

Movement

Mass C remains in equilibrium with no horizontal forces acting on it and its weight being supported by the table. It does not move.

However, Mass A starts falling toward the floor, dragging Mass B horizontally towards the pulley.

The driving force

The only force causing Mass A and Mass B to move is the weight of Mass A (49.05 newtons), which has not changed. However, that force is now trying tomove a total of 7 kg instead of only 5 kg as in the first dynamic scenario above.

A force of 49.05 newtons is not sufficient to cause a mass of 7 kg to accelerateat 9.81 m/s^2. Instead, the acceleration of each mass is proportional to the force and inversely proportional to the total mass.

a = f/m = (49.05*newtons)/(7*kg)

Entering the rightmost expression into the Google search box and pressing Enter tells us that

a = 7.0 m / s^2

Thus, the acceleration of Mass A and Mass B is 7 m/s^2, a little less than theacceleration of gravity.

What is the value of tension P?

Tension P is exerting a horizontal force on the right side of Mass B that is causing that mass to accelerate at 7.0 m / s^2. The force required to achievethat acceleration on a mass of 2 kg is

P = m*a = 2 kg*7.0 m / s^2

Once again using the Google calculator, we learn that

P = 14 newtons

Thus, although the tension at P did not go to zero when the cord was cut at Q, the resulting tension in the cord at P was substantially reduced relative tothe tension at P while the system was in equilibrium.

Third dynamic scenario

Now assume that we start over with the original static scenario and someone cuts the cord that attaches Mass C to the wall. Whathappens to the tension in each cord? Which, if any of the masses move, and if they move, what is their acceleration?