An ideal string passes over a fixed pulley and carries masses M1 = 12kg and M2 = 8kg tied at its end. Calculate the acceleration of masses and tension in the string when released (g=10m/s2)
I know the answers are
a = 2 m/s^2
T = 96 N
But I don't know how to get there
the net accelerating force is the difference in the weights (gravitational force) of the two masses
a = f / m = (120 N - 80 N) / 20 kg
find tension by looking at the smaller mass
... net force required for acceleration
T - W = 8 kg * 2 m/s^2 ... T - 80 N = 16 N
Thanks, Scott!!
To calculate the acceleration of the masses and the tension in the string, we can use Newton's second law of motion. This law states that the net force acting on an object is equal to the product of its mass and acceleration.
First, let's calculate the net force on each mass:
For M1:
The force acting on M1 is the tension in the string (T), which is pulling M1 upwards. So, the net force on M1 is given by:
F1 = T
For M2:
The force acting on M2 is the tension in the string (T), which is pulling M2 downwards. In addition, we have the gravitational force acting on M2, which is given by:
F2 = M2 * g
where g is the acceleration due to gravity (10 m/s^2).
Since the masses are connected by a string and are being pulled in opposite directions, we can set up an equation of equilibrium:
F1 = F2
T = M2 * g
Now, we can substitute the value of M2 and g to find T:
T = 8 kg * 10 m/s^2
T = 80 N
Next, we can calculate the acceleration of the masses using the net force equation:
F = m * a
For M1:
F1 = M1 * a1
T = 12 kg * a1
For M2:
F2 = M2 * a2
M2 * g = 8 kg * a2
Since the masses are connected by a string and are being pulled in opposite directions, the net force is the difference between the forces acting on the masses:
F = F2 - F1
M2 * g - T = M2 * a2 - M1 * a1
Substituting the values of M1, M2, T, and g:
8 kg * 10 m/s^2 - 80 N = 8 kg * a2 - 12 kg * a1
80 kg m/s^2 = 8 kg * a2 - 12 kg * a1
80 kg m/s^2 = 8 kg * a2 - 12 kg * a1
Dividing by 4 kg:
20 m/s^2 = 2 * a2 - 3 * a1
Since the masses are connected by a string, their accelerations are the same, albeit in opposite directions:
a1 = -a2
Substituting a1 = -a2 in the equation:
20 m/s^2 = 2 * (-a1) - 3 * a1
20 m/s^2 = -2 * a1 - 3 * a1
20 m/s^2 = -5 * a1
Dividing by -5:
a1 = -4 m/s^2
Since a1 = -a2, a2 = - (-4 m/s^2) = 4 m/s^2
Therefore, the acceleration of the masses is a = 4 m/s^2 and the tension in the string is T = 80 N.
To find the acceleration of the masses and the tension in the string, we can use Newton's second law of motion and apply it to each individual mass.
1. Begin by drawing a free-body diagram for each mass. For the mass M1 (12 kg), there are two forces acting on it: its weight (W1 = M1 * g) acting vertically downwards, and the tension in the string (T) acting upwards.
/\
/ \
/ \
/ M1 \
/________\
| T |
| |
| W |
| |
-------
2. We can now write the equations of motion for each mass:
- For M1: T - W1 = M1 * a (Equation 1)
- For M2: W2 - T = M2 * a (Equation 2)
3. Next, we need to substitute the values into the equations:
- For M1: T - (M1 * g) = M1 * a
- For M2: (M2 * g) - T = M2 * a
Now, we have two equations with two unknowns (T and a).
4. Rearrange Equation 1 to solve for T:
T = M1 * a + M1 * g
5. Substitute the value of T in Equation 2:
(M2 * g) - (M1 * a + M1 * g) = M2 * a
6. Simplify the equation:
(M2 * g) - M1 * a - M1 * g = M2 * a
7. Combine the terms:
- M1 * a - M2 * a = (M2 * g) - M1 * g
- (M1 - M2) * a = (M2 * g) - M1 * g
8. Divide both sides of the equation by (M1 - M2):
a = [(M2 * g) - (M1 * g)] / (M1 - M2)
9. Substitute the given values:
a = [(8 kg * 10 m/s^2) - (12 kg * 10 m/s^2)] / (12 kg - 8 kg)
a = (80 N - 120 N) / 4 kg
a = -40 N / 4 kg
a = -10 m/s^2
Note: The negative sign in the acceleration means that the direction of the acceleration is opposite to the assumed positive direction.
However, in this case, the negative sign indicates an error as the acceleration should not be negative. It suggests that our assumed direction for the force of tension is incorrect. We are assuming that the tension is acting in the same direction as the acceleration. In reality, the tension force is acting in the opposite direction. To resolve this, we need to reverse the direction of the tension force when calculating.
10. Let's recalculate the tension force (T) using Equation 1, but this time subtracting the weight force instead of adding it:
T = M1 * a - M1 * g
T = 12 kg * a - 12 kg * 10 m/s^2
T = 12 kg * a - 120 N
11. Substitute the value of T in Equation 2:
(M2 * g) - [12 kg * a - 120 N] = M2 * a
12. Simplify the equation:
80 N - 12 kg * a + 120 N = 8 kg * a
13. Combine like terms:
-12 kg * a + 200 N = 8 kg * a
14. Rearrange the equation to solve for a:
20 kg * a = 200 N
a = 200 N / 20 kg
a = 10 m/s^2
15. Now that we have the correct value for the acceleration, we can substitute it into Equation 1 to find T:
T = 12 kg * 10 m/s^2 - 12 kg * 10 m/s^2
T = 120 N
Therefore, the acceleration of the masses is 10 m/s^2, and the tension in the string is 120 N.