two masses m1 = 12 kg and m2 = 13 kg are connected by a light inextensible string passing over a frictionless pulley. Find (a) acceleration of the system when released from rest (b) tension in the string ?
To find the acceleration of the system, we can use Newton's second law.
Step 1: Identify the forces acting on each mass:
- For mass m1 (12 kg): It experiences its weight (mg1) downward and tension in the string (T) upward.
- For mass m2 (13 kg): It experiences its weight (mg2) downward and tension in the string (T) upward.
Step 2: Write the equations of motion for each mass:
- For mass m1: m1 * a = T - m1 * g
- For mass m2: m2 * a = m2 * g - T
Step 3: Substitute the known values into the equations:
- m1 = 12 kg
- m2 = 13 kg
- g = 9.8 m/s^2 (acceleration due to gravity)
For mass m1: 12 * a = T - 12 * 9.8
For mass m2: 13 * a = 13 * 9.8 - T
Step 4: Solve the equations simultaneously to find the acceleration and tension:
12a + T = 12 * 9.8
13a + T = 13 * 9.8
Multiply the second equation by -1:
12a + T = 12 * 9.8
-13a - T = -13 * 9.8
Add the two equations:
12a + (-13a) + T + (-T) = 12 * 9.8 + (-13 * 9.8)
-a = 117.6 - 127.4
Simplify:
-a = -9.8
Take the absolute value:
a = 9.8 m/s^2
Step 5: Substitute the value of a back into either of the original equations to find the tension:
m1 * a = T - m1 * g
12 * 9.8 = T - 12 * 9.8
Simplify:
117.6 = T - 117.6
T = 235.2 N
Therefore, the (a) acceleration of the system when released from rest is 9.8 m/s^2, and (b) the tension in the string is 235.2 N.
To find the acceleration of the system, we can apply Newton's second law of motion.
Step 1: Draw a free-body diagram for each mass separately.
For m1 (12 kg):
- There is a tension force (T) acting upwards due to the string.
- There is the weight force (m1 * g) acting downwards, where g is the acceleration due to gravity.
For m2 (13 kg):
- There is a tension force (T) acting downwards due to the string.
- There is the weight force (m2 * g) acting downwards.
Step 2: Write down the equations of motion for each mass.
For m1:
- T - m1 * g = m1 * a1 (equation 1)
Here, a1 is the acceleration of m1.
For m2:
- m2 * g - T = m2 * a2 (equation 2)
Here, a2 is the acceleration of m2.
Note that the tension force on both sides of the pulley is the same because the string is inextensible.
Step 3: Apply the constraint equation.
Since the two masses are connected by a light inextensible string passing over a frictionless pulley, the constraint equation is:
- a2 = -a1 (equation 3)
This means that the acceleration of m2 is equal in magnitude but opposite in direction to the acceleration of m1.
Step 4: Solve the equations simultaneously.
From equations 1 and 2, we can eliminate T by adding the equations:
- T - m1 * g + m2 * g - T = m1 * a1 + m2 * a2
Simplifying,
- m1 * g + m2 * g = m1 * a1 + m2 * a2
Dividing both sides by (m1 + m2),
- g = (m1 * a1 + m2 * a2) / (m1 + m2)
But from equation 3, a2 = -a1. Substituting in the above equation,
- g = (m1 * a1 - m2 * a1) / (m1 + m2)
Simplifying further,
- g = a1 * (m1 - m2) / (m1 + m2)
Rearranging,
- a1 = -g * (m1 - m2) / (m1 + m2)
Step 5: Calculate the values.
Substituting the given values:
- a1 = -9.8 m/s^2 * (12 kg - 13 kg) / (12 kg + 13 kg)
- a1 = -9.8 m/s^2 * (-1 kg) / (25 kg)
- a1 = 0.392 m/s^2
Since a2 = -a1, we have:
- a2 = -0.392 m/s^2
Therefore, the acceleration of the system when released from rest is 0.392 m/s^2.
To find the tension in the string, we can use either equation 1 or equation 2.
Using equation 1:
- T - m1 * g = m1 * a1
Substituting the given values:
- T - 12 kg * 9.8 m/s^2 = 12 kg * 0.392 m/s^2
- T - 117.6 N = 4.704 N
- T = 122.304 N
Therefore, the tension in the string is 122.304 N.
i forgot add this :
ag = 9.8 m/s^2
(a) a = F/ m
a=1kg.m/s^2/1kg