Acceleration/Tension Question?

Question

A mass, m1 = 5.00 kg, resting on a frictionless horizontal table is connected to a cable that passes over a pulley and then is fastened to a hanging mass, m2 = 9.0 kg, as in Figure P4.25. Find the acceleration of each mass and the tension in the cable.

FInd acceleration of m1 and m2 and tension in Newtons

Answer 1

In summary, (assuming acceleration due to gravity is 9.81 m/s/s)

*) Tension in Cable = 31.53 Newtons
*) Acceleration of m1 = 6.31 m/s/s
*) Acceleration of m2 = 6.31 m/s/s (the same as for m1)

Let me explain:

The second mass, m2, has the force of gravity (its weight) pulling down on it and the tension of the cable pulling up on it.

The first mass, m1, has the tension of the cable pulling to the right of it.

Call the tension in the cable F and the weight of the second block W. (note that W = m2*g, where g is the acceleration due to gravity) The NET downward force on the second block is (W-F). The net rightward force on the first block is F.

We know that the second block will accelerate down at the same rate as the first block accelerates to the right. Thus, we can equate their accelerations, which are derived by dividing their NET forces by their masses:

(W-F)/m2 = F/m1

which is:

W/m2 - F/m2 = F/m1

which leads us to:

F = (W/m2)/(1/m1 + 1/m2)

Now, remember that W=m2*g. That means that F (the tension in the cable) simplifies to (assuming g=9.81 m/s/s):

F = g/(1/m1 + 1/m2) = (9.81 m/s/s)/( (1/(5kg) + 1/(9kg)) ) = 31.53 Newtons

This is the TENSION IN THE CABLE. The acceleration of the first block is:

F/m1 = (g/m1)/(1/m1 + 1/m2) = g/(1 + m1/m2) = 6.31 m/s/s

The acceleration of the second block:

(W-F)/m2 = (m2*g - F)/m2 = g - F/m2 = g - g/(m2/m1 + 1) = g*(m2/m1)/( m2/m1 + 1 ) = g/(1 + m1/m2) = 6.31 m/s/s

As expected, the acceleration on both blocks is the same.

In fact, the acceleration is less than the acceleration due to gravity (9.81 m/s/s) due to the additional mass which is not being acted on directly by gravity. You have increased the inertia of the system without increasing the force due to gravity. Thus, the acceleration is going to be less than what you might expect.

In other words, this system has an input force equal to the weight of the second block. That force has to be distributed across both masses since the first mass is moving horizontally. This means that if the first mass is very massive, then both masses will move very slowly because the small force of the weight of the second mass will not have much ability to move both masses.

An electrical analog to this system would be the parallel combination of resistors being driven by a current source. In this case, resistance would be thought of as an analog to mass, current is an analog to gravitational acceleration, and tension in the cable is an analog to voltage. The net resistance (net mass) is thus lower. Since the current stays the same, the voltage will also be lower. In this model, if the "resistance" m1 becomes infinite (an open circuit) then all of the current goes through m2. This means that the tension in the cable will be exactly the same as m2's weight. Similarly, if the "resistance" m1 becomes 0 (a short circuit) the tension in the cable goes to 0 and m2 falls with acceleration g.

So, again, in summary:

*) Tension in Cable = 31.53 Newtons
*) Acceleration of m1 = 6.31 m/s/s
*) Acceleration of m2 = 6.31 m/s/s (the same as for m1)

Answer 2

I hope I am not doing your homework for you! First draw a "free-body" diagram (a diagram where only the external forces acting on the mass are shown) for each mass and apply Newtons Equation F=ma, were F is the vector addition of all the forces, m is the mass and a is the acceleration. Let T be the tension in the cable, g be the acceleration of gravity in the MKS system (9.8 meters per second square) and a the unknown acceleration. For the 5 kg. mass the equation becomes: T = 5a, for the 9 kg. mass the equation becomes 9g - T = 9a. Solve and you get a = (9/14)g = 6.3 meters per second square, and T = 5(9/14)g = 31.5 newtons.

Answer 3

Consider mass m1
Its weight is balanced by normal reaction
Net force acting on m1=4*9.81N
Acceleration of m1=4*9.81N/5kg=7.848m/sec^2
Mass m2 will accelerate at 9.81m/sec^2
Tension in the Rope=net force=4*9.81N=39.24N