Static Equilibrium of a Rod attached to a Pivot

In summary: The gravitational forces have no horizontal component, so that leaves only the horizontal component of the torque. But why T is not horizontal?The torque is not horizontal because the pivot is at the end of the rod that is closest to the mass.
  • #1
bbsgirl10
14
0
Homework Statement
Mrod = 0.025, m = 0.075, r1 =0.165m r2 =0.35m, T = 1.21N Angle = 40 degree
What is the formula to find horizontal and vertical components of the force exerted by the pivot on the rod?
1) Equilibrium condition to calculate the horizontal and vertical components of the force exerted by the pivot on the rod.
2) Pivot exerts a force on the rod, why was this not included in the calculation of the torques?
Relevant Equations
∑ F = 0
1665335604658.png
 
Physics news on Phys.org
  • #2
At which end of the rod is the pivot?
bbsgirl10 said:
2) Pivot exerts a force on the rod, why was this not included in the calculation of the torques?
What calculation of the torques? Please show it and include your thoughts about this problem to receive help.
 
  • Like
  • Skeptical
Likes topsquark and bbsgirl10
  • #3
kuruman said:
At which end of the rod is the pivot?

What calculation of the torques? Please show it and include your thoughts about this problem to receive help.
τ1= T r1Sinθ and τ2=( mgr2+Mrodgr1) Cos θ

1665339143701.png
 
  • #4
bbsgirl10 said:
Homework Statement:: Mrod = 0.025, m = 0.075, r1 =0.165m r2 =0.35m, T = 1.21N Angle = 40 degree
We now know where the pivot is meant to be but:
- what are the units for Mrod and m? lbs, kg, grams, metric tons, other?
- is T meant to act horizontally?
- Is r2 the distance between m and the rod's centre, or the distance between m and the pivot?

bbsgirl10 said:
What is the formula to find horizontal and vertical components of the force exerted by the pivot on the rod?
Since the system (rod & m) is in equilibriun:
vector sum of all x-forces on the system =0
vector sum of all y-forces on the system =0
It's up to you to apply these. Show us your attempt.

bbsgirl10 said:
2) Pivot exertsa force on the rod, why was this not included in the calculation of the torques?
If a force (F) acts through a point (P), what is the torque of F about P?
 
  • Like
  • Skeptical
Likes topsquark, bbsgirl10 and Lnewqban
  • #5
Steve4Physics said:
We now know where the pivot is meant to be but:
- what are the units for Mrod and m? kg
- is T meant to act horizontally?
T is a string tension
Steve4Physics said:
- Is r2 the distance between m and the rod's centre, or the distance between m and the pivot?
Distance from pivot to center hole (r1)
Distance from pivot to brass mass (r2)

Steve4Physics said:
Since the system (rod & m) is in equilibriun:
vector sum of all x-forces on the system =0
vector sum of all y-forces on the system =0
It's up to you to apply these. Show us your attempt.
i just want to know the formula to find horizontal and vertical components of the force exerted by the pivot on the rod?
i know the sum of the force will be zero but I don't get it how to calculate fx and fy
Steve4Physics said:
If a force (F) acts through a point (P), what is the torque of F about P?
 
  • #6
bbsgirl10 said:
i know the sum of the force will be zero but I don't get it how to calculate fx and fy
The sum of all forces in the x direction will be zero. Can you write down an equation from that fact?
The sum of all forces in the y direction will be zero. Can you write down an equation from that fact?
 
  • Like
  • Skeptical
Likes topsquark and bbsgirl10
  • #7
jbriggs444 said:
The sum of all forces in the x direction will be zero. Can you write down an equation from that fact?
The sum of all forces in the y direction will be zero. Can you write down an equation from that fact?
For horizontal equilibrium: ∑𝐹𝑥 = 0 (Summation of horizontal forces equal zero)
T - Fx = 0
Fx = T
where Fx= 𝑟1 cos θ
For vertical equilibrium: ∑𝐹𝑦 = 0
-m(g) - Mrod (g) + Fy = 0
Fy = mg + Mrod (g)

is this correct?
 
  • #8
bbsgirl10 said:
where Fx= 𝑟1 cos θ
r1 is a distance, not a force.
 
  • Like
  • Sad
Likes topsquark and bbsgirl10
  • #9
haruspex said:
r1 is a distance, not a force.
i mean the force acting on Fx Is tension on the string which is why I put r1 cosθ
 
  • #10
bbsgirl10 said:
i mean the force acting on Fx Is tension on the string which is why I put r1 cosθ
Doesn't help.. r1 is not a force, so the equation makes no sense. A force can only equal another force, not a distance.
I should also have pointed out that this is incorrect:
bbsgirl10 said:
T - Fx = 0
T is not horizontal, Fx is, so they cannot balance.
 
  • Like
  • Sad
Likes topsquark and bbsgirl10
  • #11
haruspex said:
Doesn't help.. r1 is not a force, so the equation makes no sense. A force can only equal another force, not a distance.
I should also have pointed out that this is incorrect:

T is not horizontal, Fx is, so they cannot balance.
Oh i see then what are the forces acting on Fx and Fy ?
 
  • #12
bbsgirl10 said:
Oh i see then what are the forces acting on Fx and Fy ?
Fx is a horizontal force. It needs to be balanced by a horizontal force, or horizontal component of a force. The gravitational forces have no horizontal component, so that leaves only the horizontal component of …?
 
  • Like
Likes topsquark and bbsgirl10
  • #13
haruspex said:
Fx is a horizontal force. It needs to be balanced by a horizontal force, or horizontal component of a force. The gravitational forces have no horizontal component, so that leaves only the horizontal component of …?
1665386452140.png

But why T is not horizontal?
 
  • #14
haruspex said:
Fx is a horizontal force. It needs to be balanced by a horizontal force, or horizontal component of a force. The gravitational forces have no horizontal component, so that leaves only the horizontal component of …?
Tension ?
 
Last edited:
  • #16
:nb)
 
  • #17
topsquark said:
Look at your problem statement. What is ##\theta##?

-Dan
Theta refers to the angle measured by the sensor minus 90 degree which is 40 degree.
I m so lost right now what are the factors involved in Fx and Fy? If the sum is zero do I assume the factors that affect also zero?
 
  • #18
bbsgirl10 said:
Theta refers to the angle measured by the sensor minus 90 degree which is 40 degree.
I m so lost right now what are the factors involved in Fx and Fy? If the sum is zero do I assume the factors that affect also zero?
What sensor? We are apparently missing information and we can't help you properly until we know what's going on. Can you please post the whole problem statement? Strings, sensors, pivot points, and anything else that you may not have mentioned.

-Dan
 
  • Like
Likes Steve4Physics
  • #19
topsquark said:
What sensor? We are apparently missing information and we can't help you properly until we know what's going on. Can you please post the whole problem statement? Strings, sensors, pivot points, and anything else that you may not have mentioned.

-Dan

Mrod = 0.025kg, m = 0.075kg, r1 =0.165m r2 =0.35m, T (String Tension) = 1.21N, Angle Theta = 40 degree.
Distance from pivot to center hole (r1)
Distance from pivot to brass mass (r2)
1) Equilibrium condition to calculate the horizontal and vertical components of the force exerted by the pivot on the rod.
 
  • #20
θ is the angle between the rod and T. Although the diagram makes it look like T is horizontal, this will only be true if the angle between the rod and the horizontal is also θ (=40º). This is not specified.

Let α be the angle between the rod and horizontal. We can calculate α by taking moments about the pivot using the given data.

When I calculate α I get α = 64.5º. Not 40º.

So T is not horizontal. But maybe someone else would like to check.

EDIT: Changed in the light of the extra information in Post #19.
 
  • Like
Likes topsquark
  • #21
bbsgirl10 said:
Tension ?
Yes. So if the tension is T, what is its horizontal component?
 
  • Like
Likes topsquark
  • #22
Steve4Physics said:
θ is the angle between the rod and T. Although the diagram makes it look like T is horizontal, this will only be true if the angle between the rod and the horizontal is also θ (=40º). This is not specified.

Let α be the angle between the rod and horizontal. We can calculate α by taking moments about the pivot using the given data.

When I calculate α I get α = 64.5º. Not 40º.

So T is not horizontal. But maybe someone else would like to check.

EDIT: Changed in the light of the extra information in Post #19.
I got α = 64.5º for the angle between the rod and the horizontal under the assumption that the tension is horizontal and has magnitude 1.21 N.
 
  • Like
Likes topsquark
  • #23
kuruman said:
I got α = 64.5º for the angle between the rod and the horizontal under the assumption that the tension is horizontal and has magnitude 1.21 N.
The problem with that is ##\alpha= 64.5^o## and ##\theta = 40^o## are not consistent with T being horizontal.

The clockwise moment produced by the tension is ##T r_1 \sin(40^o)## but this doesn’t rely on T being horizontal.
 
  • Like
Likes topsquark
  • #24
Steve4Physics said:
The problem with that is ##\alpha= 64.5^o## and ##\theta = 40^o## are not consistent with T being horizontal.

The clockwise moment produced by the tension is ##T r_1 \sin(40^o)## but this doesn’t rely on T being horizontal.
The 40° angle is ignored in my calculation. I thought that's what you did when you got the 64.5°. We don't have a precise statement of the problem including this (presumably measured) 40° angle. All we have is
bbsgirl10 said:
Theta refers to the angle measured by the sensor minus 90 degree which is 40 degree.
If it is the angle between the tension vector and the rod as shown in the drawing, then the angle between between the rod and the horizontal (by my calculation) is 72.2°. I don't feel confident that I understand what's what with this problem.
 
  • Like
Likes topsquark and Lnewqban
  • #25
kuruman said:
The 40° angle is ignored in my calculation. I thought that's what you did when you got the 64.5°. We don't have a precise statement of the problem including this (presumably measured) 40° angle. All we have is

If it is the angle between the tension vector and the rod as shown in the drawing, then the angle between between the rod and the horizontal (by my calculation) is 72.2°. I don't feel confident that I understand what's what with this problem.
Yes, the problem is very badly posed. We can:
- assume that T is horizontal and not use θ=40° or
- assume that θ=40º and treat T as non-horizontal.
Unless the OP can explain, the problem won't get resolved.
 
  • Like
Likes topsquark, Lnewqban and kuruman

1. What is static equilibrium?

Static equilibrium refers to the state of an object when it is at rest and the net force acting on it is zero. This means that the object is not moving and all the forces acting on it are balanced.

2. How is static equilibrium achieved in a rod attached to a pivot?

In a rod attached to a pivot, static equilibrium is achieved when the forces acting on the rod are balanced and the rod is not rotating. This means that the sum of all the forces acting on the rod must be equal to zero and the sum of all the torques acting on the rod must also be equal to zero.

3. What is the role of the pivot in static equilibrium?

The pivot acts as a fixed point of rotation for the rod. This means that the rod can rotate freely around the pivot without any external forces acting on it. The pivot also helps to distribute the forces and torques acting on the rod evenly, allowing for static equilibrium to be achieved.

4. How does the position of the pivot affect static equilibrium in a rod?

The position of the pivot can affect the distribution of forces and torques acting on the rod. If the pivot is closer to one end of the rod, the forces and torques acting on that end will be greater, which may result in the rod not being in static equilibrium. The position of the pivot must be carefully chosen to ensure that the forces and torques acting on the rod are balanced.

5. What are some real-life examples of static equilibrium in a rod attached to a pivot?

Some examples of static equilibrium in a rod attached to a pivot include a seesaw, a balance scale, and a door on its hinges. In all of these examples, the pivot acts as the fixed point of rotation for the rod, allowing it to remain in static equilibrium.

Similar threads

  • Introductory Physics Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
386
  • Introductory Physics Homework Help
Replies
13
Views
1K
  • Introductory Physics Homework Help
Replies
14
Views
992
  • Introductory Physics Homework Help
Replies
3
Views
245
  • Introductory Physics Homework Help
Replies
12
Views
617
  • Introductory Physics Homework Help
Replies
14
Views
631
  • Introductory Physics Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
5
Views
1K
Back
Top