I was given a systems equation. Two masses are on a flat frictionless surface tied together. There is a pulley type thing and another mass is hanging over the edge of the table. It is attatched to the other 2 masses. It has no Fn, only Fg because it is hanging.
So I draw lines of Fg and Fn on the other masses, but they cancel out.
M1=12g
M2=8g
I must find out what T1 and T2 are?
Which are the tensions of the rope/string between each of the masses. The problem says everything is frictionless. But then it says the objects are released from rest. (what does that mean?)
When the problem states that the objects are released from rest, it means that initially, all the masses are stationary and not moving.
To find the tensions T1 and T2 in the rope/string between each of the masses, we can analyze the forces acting on the system:
1. Mass M1 (12g):
- It experiences a gravitational force (Fg1 = m1 * g), where m1 is the mass of M1 and g is the acceleration due to gravity.
- It is connected to the pulley system, so it also experiences tension T1 pulling it to the right.
2. Mass M2 (8g):
- It experiences a gravitational force (Fg2 = m2 * g), where m2 is the mass of M2 and g is the acceleration due to gravity.
- It is connected to the pulley system, so it also experiences tension T1 pulling it to the left.
- It is also connected to the hanging mass, so it experiences tension T2 pulling it upward.
3. Hanging mass:
- Since the hanging mass is only influenced by gravitational force (Fg3 = m3 * g), and not any normal force, we can assume it experiences tension T2 pulling it downward.
Now, let's apply Newton's second law (F = ma) to each of the masses:
1. Mass M1:
Sum of forces in the x-direction: T1 - Fg1 = m1 * a1 (Since there is no acceleration in the x-direction, this equation becomes T1 = Fg1)
2. Mass M2:
Sum of forces in the x-direction: T1 - T2 - Fg2 = m2 * a2 (Since there is no acceleration in the x-direction, this equation becomes T1 - T2 = Fg2)
3. Hanging mass:
Sum of forces in the y-direction: T2 - Fg3 = m3 * a3 (Since there is no acceleration in the y-direction, this equation becomes T2 = Fg3)
Now, substitute the given masses m1 = 12g and m2 = 8g, and acceleration due to gravity g into the equations to solve for T1 and T2.
To solve this problem, let's break it down step by step:
1. Start by drawing a free body diagram for each mass. In this case, we have three masses: M1, M2, and the hanging mass. Label the forces. Since everything is frictionless, the only forces acting on each mass are the force of gravity (Fg) and the tension forces (T1 and T2).
2. The hanging mass experiences only the force of gravity since it is not on a surface. So, draw a downward arrow to represent Fg on the hanging mass.
3. For the other two masses, draw both Fg and Fn arrows. The force of gravity (Fg) is downward, and the normal force (Fn) is upward. Since the surfaces are frictionless, Fn is equal in magnitude and opposite in direction to Fg.
4. Now, using Newton's second law (F = ma), write the equations of motion for each mass. Since all the masses are connected by the same rope or string, their accelerations are the same. Let's assume the acceleration is "a".
- For M1: T1 - M1g = M1a (1)
- For M2: T2 - M2g = M2a (2)
- For the hanging mass: Mhg - Mhg = Mhga (3) [since Fg and Fn cancel each other out]
5. The hanging mass is released from rest, which means it initially has zero velocity. Therefore, its initial acceleration is also zero (a = 0).
6. Now, we can solve the system of equations. Since a = 0, equation (3) becomes Mhg - Mhg = 0, which simplifies to 0 = 0. This equation does not provide any new information.
7. Solving equations (1) and (2) simultaneously, we eliminate a to find the values of T1 and T2. Substitute the value of a from equation (2) into equation (1):
T1 - M1g = M1(M2g + M1g) / (M1 + M2)
Simplify the equation and plug in the given values for M1 and M2:
T1 - 12g = (12g)(8g + 12g) / (12g + 8g)
T1 - 12g = (12g)(20g) / (20g)
T1 - 12g = 240g
T1 = 252g
Therefore, T1 = 252g.
8. Now, substitute the value of T1 back into equation (2):
252g - 8g = 8g(8g) / (12g)
Simplify the equation and plug in the given values:
244g = 64g^2 / 12g
244g = 64g
Divide both sides by g:
244 = 64
This is not a valid solution, and it means that there might be an error in the problem statement or calculations. Please recheck the given values or review the problem setup.
To summarize, in a system where two masses are connected to a hanging mass through a frictionless pulley, you can determine the tensions T1 and T2 by applying Newton's second law and solving the system of equations. However, in this case, the provided values lead to an inconsistent result, suggesting a problem with the setup or calculations.