Torque on two pillars (introductory physics problem)

[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.

 

[BvU]

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 ?

$\$

 

[haruspex]

since they are spaced an equal distance apart

from what?

 

[Camden]

from what?

From each other

 

[Camden]

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.

 

[haruspex]

From each other

How can two objects not be at the same distance from each other??

 

[BvU]

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) ?

$\$

 

[haruspex]

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.

 

[Camden]

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.

 

[Camden]

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.

 

[Lnewqban]

... 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.

Please, see:
https://courses.lumenlearning.com/a...ter/9-2-the-second-condition-for-equilibrium/

https://courses.lumenlearning.com/a...statics-including-problem-solving-strategies/

 

[haruspex]

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.

 

[haruspex]

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?

 

[Camden]

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 ?

 

[haruspex]

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.