F = MA 2010 # 11 (Static equilibrium, 3 masses on a string)

In summary: I can't do the problem for you, but I can give hints and help you along the way. Have you drawn a free-body diagram yet? That is always the first step in any mechanics problem. Once you have that, you can start writing equations for the forces in each direction. Remember, the sum of all the forces in any direction must be zero, according to Newton's Second Law. So for the horizontal forces, you should have T1x + T2x = 0. What can you say about the vertical forces?In summary, the conversation is discussing question 11 from a physics homework assignment. The problem involves finding the tension in two strings supporting a central mass, and the conversation covers
  • #1
SignaturePF
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Homework Statement


See:
http://www.aapt.org/physicsteam/2010/upload/2010_FmaSolutions.pdf
Question 11

Homework Equations


T_net = 0
F_net = 0

The Attempt at a Solution


Start with F_net = 0:
Let T1 be the force of tension on the left, T2 on the right
T1cosθ=T2cosθ
Y components:
T1sinθ+T2sinθ = mg
Since T1 = T2, as established by the x-component force equation:
2T1sinθ= mg
T1sinθ = mg /2

Now let's look at T_net = 0
Take the leftmost point to be the PoR
This is where I got stuck.
T1cosθb = T2sinθa
T2sinθ=T1sinθ = mg/2
T1cosθb = mg/2a
a/b = mg/2T1cosθ

I'm really confused about how to find the torque:
I get that it is rxF or rFsintheta but I'm struggling to make the equation.
 
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  • #2
There is no tension in question 12 at that link.
 
  • #3
Sorry I meant question 11.
 
  • #4
SignaturePF said:
Let T1 be the force of tension on the left, T2 on the right
T1cosθ=T2cosθ
What is [itex]\vartheta[/itex]? If you are assuming that the angle on each side is the same, you have to justify that.
I'm really confused about how to find the torque:
You don't need torque in this problem.
 
  • #5
Huh? How would you do the problem then?
Ok, instead of justifying that, I'll just split it into components.
T1x = T2x
I'm not at all sure how to proceed. Could you give me a hint?
 
  • #6
Start with a free-body diagram. Look at the leftmost mass. What are the forces on it? What are the forces on the other masses? What can you say about the tensions in the two strings?

The next step is, as you say, breaking the forces on the center mass into components.
 
  • #7
Left-most mass:
Down: mg
Up: T0

Middle Mass:
Left: T1x
Right: T2x
Up: T1y, T2y
Down: mg

Right-most mass:
Down: mg
Up: T3

Thus we know T1x = T2x and that T1y + T2y = mg
Also T0 = T3 = mg
so T1y + T2y = T0 = T3

How could we relate this to a,b?
 
  • #8
SignaturePF said:
so T1y + T2y = T0 = T3
Up to here everything is okay, and [itex]T_0 = T_3[/itex] is okay, too, [strike]but the first term is a problem[/strike]. If the tension at one end of a string is [itex]T[/itex], what is the tension at the other end?
How could we relate this to a,b?
Trigonometry.
 
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  • #9
I don't see why T1y + T2y = mg is wrong.
The two y-components of the two tensions need to add up to equal the weight of the middle sphere.

Using the trig:
T2sinθ=b
T1sinθ2 =a

??
 
  • #10
SignaturePF said:
I don't see why T1y + T2y = mg is wrong.
It isn't. I don't know what I was thinking.
Using the trig:
T2sinθ=b
T1sinθ2 =a

??
By now you know that all the tensions are equal, so you can drop the subscripts.

Next you should look at the horizontal components of the forces on the central mass.
 
  • #11
We know that 2Tcosθ = mg
This implies that Tcosθ = .5mg
So θ is logically 60 degrees.
I'm still stuck. Could you just show me?
 
  • #12
SignaturePF said:
We know that 2Tcosθ = mg
You are again assuming that the angles are equal. They are, but you have to demonstrate it first; you can't just assume it.

Which angle is θ? Between the string and the horizontal, or between the string and the vertical? You should look at the horizontal forces first; that will be a big help.
Could you just show me?
That's against the rules of PF.
 

FAQ: F = MA 2010 # 11 (Static equilibrium, 3 masses on a string)

What is F = MA 2010 #11 about?

F = MA 2010 #11 is a problem that deals with static equilibrium, specifically involving three masses connected by a string. It is a common physics problem that tests a student's understanding of forces and how they interact in a static system.

What is static equilibrium?

Static equilibrium is a state in which all forces acting on an object are balanced, resulting in zero net force and zero acceleration. This means that the object is at rest or moving at a constant velocity.

How do you solve for static equilibrium in this problem?

To solve for static equilibrium in this problem, you must first draw a free body diagram and identify all the forces acting on each of the three masses. Then, you can use the equations of static equilibrium, such as ΣF = 0 and Στ = 0, to set up and solve a system of equations to find the unknown forces and tensions in the system.

What is the significance of F = MA 2010 #11 in physics?

F = MA 2010 #11 is a common problem that tests a student's understanding of Newton's laws of motion, specifically the relationship between force, mass, and acceleration. It also highlights the importance of understanding static equilibrium in various systems.

What are some real-life applications of F = MA 2010 #11?

F = MA 2010 #11 has practical applications in engineering, architecture, and mechanics. For example, it can be used to analyze the forces acting on a bridge or a building to ensure it is in static equilibrium and can withstand external forces. It can also be applied in designing pulley systems or cranes, where multiple masses are connected by strings or cables.

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