Statics/Equilibrium - Find the Tension in the Wire

In summary, The problem involves a horizontal bar hinged at one end and supported by a wire at the other end. A block of weight 140 N can be moved along the bar and the distance from the wall to its center of mass is 1.06 m. The tension in the wire is found using the torque relation and is equal to 116.7265 N. To find the horizontal and vertical components of the force on the bar, three equations are needed, including the vertical force at point A and the force at point B along the wire. It is important to include all forces acting on the body, even if they seem negligible. Newton's 3rd Law still applies to accelerating bodies, so it should not be disregarded
  • #1
prosteve037
110
3

Homework Statement


I've already solved the problem, though I didn't understand WHY I took the steps I did; I just want to know why this is the way to solving the problem. Here's the question:

In the figure below, a thin horizontal bar AB of negligible weight and length L = 1.9 m is hinged to a vertical wall at A and supported at B by a thin wire BC that makes an angle θ = 42° with the horizontal. A block of weight W = 140 N can be moved anywhere along the bar; its position is defined by the distance x = 1.06 m from the wall to its center of mass. Find (a) the tension in the wire, and the (b) horizontal and (c) vertical components of the force on the bar from the hinge at A


Homework Equations


[itex]\sum{F}_{x}\textit{ = ma}_{x}\textit{ = 0}[/itex]

[itex]\sum{F}_{y}\textit{ = ma}_{y}\textit{ = 0}[/itex]

[itex]\sum{\tau}\textit{ = 0}[/itex]

The Attempt at a Solution



a.) Tension in the wire, T :

[itex]\sum{\tau}\textit{ = 0 = TsinθL - W}_{block}\textit{x}[/itex]

[itex]\textit{T = }\frac{W_{block}x}{sinθL}[/itex]

[itex]\textit{T = }\frac{(140 N)(1.06 m)}{sin(42°)(1.9 m)}[/itex]

[itex]\textit{T = 116.7265 N}[/itex]


Now b.) and c.) can be found easily using the other force relations.


My question is, why can't you use the force relations instead of the torque relation? I can see that it gives the wrong answer, but WHY does it give the wrong answer if you don't use the torque relation?

Is it because the weight of the bar is neglected? If so, how does that affect what you can or can't use in a problem like this?

Thanks
 
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  • #2
My question is, why can't you use the force relations instead of the torque relation? I can see that it gives the wrong answer, but WHY does it give the wrong answer if you don't use the torque relation?
It doesn't give the wrong answer. To solve part b and c, you use the force relationship. When you look at the system, you have 3 unknown external reaction forces..the x and y forces at A, and the force at B directed along the wire..so you need 3 equations to solve the problem. Don't forget the reaction at B...maybe that's what you forgot when you say you got the wrong answer. The vert force at B plus the vert force at A must equal W. The horiz force at A is balanced by the horiz force at B.
 
  • #3
Just realized I forgot to provide the figure :P Here it is:

http://edugen.wileyplus.com/edugen/courses/crs4957/art/qb/qu/c12/fig12_40.gif



PhanthomJay said:
It doesn't give the wrong answer. To solve part b and c, you use the force relationship. When you look at the system, you have 3 unknown external reaction forces..the x and y forces at A, and the force at B directed along the wire..so you need 3 equations to solve the problem. Don't forget the reaction at B...maybe that's what you forgot when you say you got the wrong answer. The vert force at B plus the vert force at A must equal W. The horiz force at A is balanced by the horiz force at B.

Ohhh so then I think I set up the force equations wrong. I think I forgot to include the components of the contact force from the wall, thinking that they were just negligible 3rd Law force pairs :P


So then here are my 2 revised force equations for this problem:

[itex]\textit{F}_{y}\textit{ + Tsinθ = W}_{block}[/itex]

[itex]\textit{Tcosθ = F}_{x}[/itex]

Look good?


I keep making the elementary mistake of excluding forces that are acting on the body of interest, thinking that Newton's 3rd Law would cancel them out. That said, should I just always focus on the forces acting ON a body and disregard the 3rd Law? That is, for problems where the system as a whole isn't accelerating?
 
  • #4
prosteve037 said:
Just realized I forgot to provide the figure :P Here it is:

http://edugen.wileyplus.com/edugen/courses/crs4957/art/qb/qu/c12/fig12_40.gif





Ohhh so then I think I set up the force equations wrong. I think I forgot to include the components of the contact force from the wall, thinking that they were just negligible 3rd Law force pairs :P


So then here are my 2 revised force equations for this problem:

[itex]\textit{F}_{y}\textit{ + Tsinθ = W}_{block}[/itex]
Yes, which direction?
[itex]\textit{Tcosθ = F}_{x}[/itex]
yes, which direction?
I keep making the elementary mistake of excluding forces that are acting on the body of interest, thinking that Newton's 3rd Law would cancel them out. That said, should I just always focus on the forces acting ON a body
yes
and disregard the 3rd Law?
I wouldn't disregard it completely, it comes in handy when looking at forces on different bodies
That is, for problems where the system as a whole isn't accelerating?
For accelerating bodies, the process is the same: look at the forces acting on the body. But Newton's 3rd Law applies to accelerating bodies as well as to non- accelerating bodies.
 
  • #5


As a scientist, it is important to understand the principles and concepts behind problem solving rather than just memorizing equations. In this problem, the key concept is equilibrium, which means that the net force and net torque acting on the system must be equal to zero. This is why we use the equations \sum{F}_{x}\textit{ = ma}_{x}\textit{ = 0} and \sum{F}_{y}\textit{ = ma}_{y}\textit{ = 0} to find the horizontal and vertical components of the force on the bar.

However, in this specific problem, we also have to consider the rotational motion of the bar around the hinge at point A. This is where the torque equation \sum{\tau}\textit{ = 0} comes into play. The torque exerted by the tension in the wire must be equal and opposite to the torque exerted by the weight of the block. This is why we use the torque equation to find the tension in the wire.

If we only use the force equations, we are not taking into account the rotational motion of the bar and therefore, we will not get the correct answer. Neglecting the weight of the bar would also affect the problem as it changes the distribution of forces and torques in the system.

In summary, it is important to understand the underlying principles and concepts when solving physics problems, rather than just relying on equations. In this case, the concept of equilibrium and the consideration of rotational motion are crucial in finding the correct solution.
 

1. How do I find the tension in a wire using statics/equilibrium?

To find the tension in a wire using statics/equilibrium, you will need the forces acting on the wire and the distance between them. Set up a free body diagram and use the equation T=mg+ma, where T is the tension, m is the mass, g is the acceleration due to gravity, and a is the acceleration. Solve for T to find the tension in the wire.

2. What are the units for tension in statics/equilibrium?

The units for tension in statics/equilibrium are Newtons (N). This is a unit of force, and it is equal to 1 kg*m/s^2. In the equation T=mg+ma, the units for T would be kg*m/s^2, which is equivalent to N.

3. Can the tension in a wire ever be negative?

No, the tension in a wire cannot be negative. Tension is a force that is always directed away from the object it is acting on. If the tension were negative, it would mean the force is acting in the opposite direction, which is not physically possible.

4. What is the difference between tension and compression?

Tension and compression are both types of forces that act on objects. Tension is a pulling force that stretches or elongates an object, while compression is a pushing force that shortens or compresses an object. In the context of a wire, tension would be the force pulling on the ends of the wire, while compression would be the force pushing on the sides of the wire.

5. What factors can affect the tension in a wire?

The tension in a wire can be affected by several factors, including the weight of the object suspended from the wire, the length of the wire, and the material and thickness of the wire. Other external forces, such as wind or friction, can also affect the tension in a wire. Additionally, any changes in the surrounding environment, such as temperature or humidity, can also impact the tension in a wire.

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