# Torque on two pillars (introductory physics problem)

• Camden
In summary, the conversation involves determining the downward force applied to each pillar of a bridge and the tricky aspect of torque coming into play. The formula for moment balance is discussed, with clarification on the terms F, r, and theta in the equation T=Fsin(theta)r. The next step involves choosing the body and axis for applying the formula.
Camden
Homework Statement
A bridge section of mass 2400kg is supported at each end of its 24.0m length by pillars. A car of mass 960kg is parked 5.00m from one end of the bridge.
Relevant Equations
F=ma
T=Fsin(theta)r
One image is attached is the question and the other is my attempt. I believe the pillars would have a downward force of 11760N applied to each pillar since they are spaced an equal distance apart. Now the tricky bit is when Torque comes into play. I believe I need to find the distance the car sits from the centre of the bridges mass. I am unsure how to do that though.

#### Attachments

• IMG_1600.jpg
21.7 KB · Views: 70
• IMG_1601.jpg
25.6 KB · Views: 71
Camden said:
I believe the pillars would have a downward force of 11760N applied to each pillar since they are spaced an equal distance apart.
My neck hurts, perhaps because the pictures are rotated ?

Your belief is not justified. If the car is on top of one of the pillars, would its weight be distributed equally over the two ?

##\ ##

Lnewqban
Camden said:
since they are spaced an equal distance apart
from what?

haruspex said:
from what?
From each other

BvU said:
My neck hurts, perhaps because the pictures are rotated ?

Your belief is not justified. If the car is on top of one of the pillars, would its weight be distributed equally over the two ?

##\ ##
I believe the pillar the car is almost directly underneath will experience more weight from the car.

Camden said:
From each other
How can two objects not be at the same distance from each other??

Much better. Any useful formula for the fraction (perhaps derived from an applicable condition for equilibrium of the horizontal section that rests on the pillars) ?

## \ ##

Camden said:
I believe the pillar the car is almost directly underneath will experience more weight from the car.
The car is above the pillars, not underneath them.
But yes, the pillar nearer the car will experience the greater load.
Use moment balance to find out how much.

haruspex said:
The car is above the pillars, not underneath them.
But yes, the pillar nearer the car will experience the greater load.
Use moment balance to find out how much.
Are there any other phrases that mean the same as Moment balance? I have not been taught that particular word before.

BvU said:
Much better. Any useful formula for the fraction (perhaps derived from an applicable condition for equilibrium of the horizontal section that rests on the pillars) ?

## \ ##
So I believe the net Torque should be zero. I am unsure which formula I should be deriving from.

Camden said:
Are there any other phrases that mean the same as Moment balance? I have not been taught that particular word before.
Moment is another word for torque, and balance means the net value is zero.

Camden said:
So I believe the net Torque should be zero. I am unsure which formula I should be deriving from.
You quoted T=Fsin(theta)r. Do you understand what F, r and θ are in that equation?

haruspex said:
You quoted T=Fsin(theta)r. Do you understand what F, r and θ are in that equation?
Yes I do,
F= Force applied
r= Distance force is being applied from
theta= angle at which the force is being applied
So is my first step 0= Fsin(theta)r ?

Camden said:
Yes I do,
F= Force applied
r= Distance force is being applied from
theta= angle at which the force is being applied
So is my first step 0= Fsin(theta)r ?
Yes, but you need to choose the body you are applying it to and the axis about which you are taking moments.
And to clarify, theta is the angle between the line of action of the force and the line along which you are measuring r.

## 1. What is torque?

Torque is a measure of the force that can cause an object to rotate around an axis. It is calculated by multiplying the force applied to an object by the distance from the axis of rotation to the point where the force is applied.

## 2. How is torque related to two pillars?

In this introductory physics problem, torque is used to determine the stability of an object placed on two pillars. The torque created by the weight of the object must be balanced by the torque created by the pillars in order for the object to remain in equilibrium.

## 3. What factors affect the torque on an object on two pillars?

The distance between the pillars, the weight of the object, and the location of the object's center of mass all affect the torque on an object on two pillars. The farther the object is from the pillars, the greater the torque will be.

## 4. How can torque on two pillars be calculated?

Torque on two pillars can be calculated by multiplying the weight of the object by the distance between the pillars and the distance from the pillars to the object's center of mass. This will give the torque created by the object. The pillars must then create an equal and opposite torque to keep the object in equilibrium.

## 5. What are some real-life examples of torque on two pillars?

Some real-life examples of torque on two pillars include balancing a ladder against a wall, a see-saw, and a shelf supported by two brackets. In each of these cases, the pillars (or wall) must create an equal and opposite torque to balance the weight of the object and keep it from falling over.

• Introductory Physics Homework Help
Replies
6
Views
746
• Introductory Physics Homework Help
Replies
7
Views
822
• Introductory Physics Homework Help
Replies
45
Views
2K
• Introductory Physics Homework Help
Replies
3
Views
181
• Introductory Physics Homework Help
Replies
4
Views
2K
• Introductory Physics Homework Help
Replies
8
Views
2K
• Introductory Physics Homework Help
Replies
28
Views
2K
• Introductory Physics Homework Help
Replies
17
Views
4K
• Introductory Physics Homework Help
Replies
2
Views
1K
• Introductory Physics Homework Help
Replies
16
Views
2K