Three blocks are located on a horizontal frictionless table. They are connected by a massless cord, as shown in the figure, and pulled to the right. The masses of the three blocks are m1 = 22 kg, m2 = 21 kg, and m3 = 33 kg. The pulling force is equal to T3 = 53 N. What is the tension T2? I tried doing the acceleration times mass (21)and I got 14.6, which is apparently wrong. What is the right answer?
Once again you did not give me the order of the blocks. However do it the same way as the other one but this is easier because gravity is not part of the problem.
To find the tension T2, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration.
In this case, we first need to find the acceleration of the system. To do that, we need to analyze the forces acting on each block.
For block m1 (22 kg):
- The tension T1 is acting to the left.
- The tension T2 is acting to the right.
- The net force on m1 is T2 - T1.
For block m2 (21 kg):
- The tension T2 is acting to the left.
- The tension T3 is acting to the right.
- The net force on m2 is T2 - T3.
For block m3 (33 kg):
- The tension T3 is acting to the left.
- The gravitational force mg is acting to the right.
- The net force on m3 is T3 - mg.
Since the system is connected and there is no friction, the acceleration of all blocks will be the same.
Now, we can set up equations for the net force on each block:
m1 * a = T2 - T1 (Equation 1)
m2 * a = T2 - T3 (Equation 2)
m3 * a = T3 - mg (Equation 3)
To solve these equations, we need one more equation. Since the pulling force T3 is given as 53 N, we can set up the following equation:
T3 = 53 N
Substituting this into Equation 3, we get:
m3 * a = 53 N - mg (Equation 4)
Now, we can solve this system of equations.
From Equation 1, we get:
T2 - T1 = m1 * a
From Equation 2, we have:
T2 - T3 = m2 * a
Substituting the given value of T3 (53 N), we get:
T2 - 53 N = m2 * a
Rearranging this equation, we have:
T2 = m2 * a + 53 N
Similarly, substituting Equation 4 into this equation, we get:
T2 = m2 * a + 53 N = m2 * a + 53 N = m3 * a + mg
Now, we can substitute the given masses and known acceleration due to gravity (g = 9.8 m/s^2) into this equation:
T2 = (21 kg) * a + 53 N = (33 kg) * a + (33 kg * 9.8 m/s^2)
Simplifying, we get:
T2 = 21a + 53 = 33a + 323.4
Rearranging this equation, we get:
12a = 270.4
Dividing both sides by 12, we find:
a = 22.53 m/s^2
Now, we can substitute this value of acceleration back into Equation 1 to find T2:
T2 - T1 = m1 * a
T2 - T1 = (22 kg) * (22.53 m/s^2)
T2 - T1 = 494.66 N
Since the system is frictionless, the tension T1 is equal to the pulling force T3:
T1 = T3 = 53 N
Substituting this value into the equation above, we get:
T2 - 53 N = 494.66 N
Rearranging, we find:
T2 = 547.66 N
Therefore, the tension T2 is approximately equal to 547.66 N.