A pulley is suspended by a Cord (C). on one end of the pulley there is a 1.2 kg block and on the other is a 3.2 kg block.Determine the tension in this cord after the masses are released and before one hits the ground. Ignore the mass of the pulley and cords
34N
if you use the formula as such as
ft=(2*m1*m2/m1+m2)2g
you will get the right answer which is 34.210
Well, well, well, looks like we've got ourselves a tension-filled situation here! Let's see if I can pull some answers out of my imaginary hat.
Since there's no friction in this scenario, the tension in the cord will be the same on both sides. So, we start by calculating the gravitational force acting on each block separately.
For the 1.2 kg block, the gravitational force would be:
F1 = m1 * g
= 1.2 kg * 9.8 m/s²
= 11.76 N
For the 3.2 kg block, we'd have:
F2 = m2 * g
= 3.2 kg * 9.8 m/s²
= 31.36 N
Now, since both blocks are attached to the same cord and pulley, the tension in the cord is the same on both sides. Therefore, we can set up an equation:
T - F1 = F2
Solving for T, we have:
T = F1 + F2
= 11.76 N + 31.36 N
= 43.12 N
So, drumroll, please... the tension in the cord before either of those blocks hits the ground is 43.12 Newtons!
Now, if only we could come up with a joke to lighten the tension in this situation...
To determine the tension in the cord after the masses are released, we need to consider the forces acting on each block.
Let's start by looking at the 1.2 kg block. It is being pulled downward by the force of gravity (weight), which we can calculate using the equation:
Weight = mass * acceleration due to gravity
Weight_1.2kg = 1.2 kg * 9.8 m/s^2
Weight_1.2kg = 11.76 N
Since the 1.2 kg block is connected to the pulley by the cord, the tension in the cord is equal to the weight of the 1.2 kg block:
Tension_1.2kg = Weight_1.2kg = 11.76 N
Now let's consider the 3.2 kg block. It is also being pulled downward by the force of gravity:
Weight_3.2kg = 3.2 kg * 9.8 m/s^2
Weight_3.2kg = 31.36 N
Since the 3.2 kg block is connected to the pulley by the same cord, the tension in the cord is equal to the weight of the 3.2 kg block:
Tension_3.2kg = Weight_3.2kg = 31.36 N
However, the tension in the cord is also affected by the motion of the system. As the masses are released, the 1.2 kg block accelerates downward while the 3.2 kg block accelerates upward. Due to the pulley system, the accelerations of both blocks will be the same.
To find the acceleration, we can use Newton's second law of motion:
Sum of forces = mass * acceleration
For the 1.2 kg block, the net force is the difference between the tension and the weight:
Sum of forces_1.2kg = Tension_1.2kg - Weight_1.2kg
For the 3.2 kg block, the net force is the sum of the tension and the weight:
Sum of forces_3.2kg = Tension_3.2kg + Weight_3.2kg
Since the net force is the same for both blocks (equal to mass times acceleration), we can set these two equations equal to each other:
Tension_1.2kg - Weight_1.2kg = Tension_3.2kg + Weight_3.2kg
Plugging in the known values:
11.76 N - 11.76 N = Tension_3.2kg + 31.36 N
Simplifying:
0 N = Tension_3.2kg + 31.36 N
Tension_3.2kg = -31.36 N
Since tension cannot be negative, this means there is no tension in the cord when the system reaches equilibrium. Therefore, the answer is 0 N.
Please note that if you release the masses without accounting for any friction or air resistance, this is the theoretical result. In practice, there might be some slight tension in the cord due to factors like friction.
Solve these two simultaneous equations with unknowns a (acceleration) and T (tension).
M1*g - T = M1*a
T - M2*g = M2*a
This leads to
a = [(M1 -M2)/(M1 +M2)]*g
and
T = [2*M1*M2/(M1+M2)]*g
I did the Ft=g[(2Mm)/(M+m)] equation and ended up with 17.1N
Which plugged in was
Ft=9.8[((2)(1.2)(3.2))/(1.2+3.2)]