Raising a 20m Ladder with Hydraulic Force: Calculating Magnitude at B

[MAPgirl23]

A ladder of length L = 20.0 m is carried by a fire truck. The ladder has a weight of 2880 N and its center of gravity is at its center. The ladder is pivoted at one end (A) about a pin; you can ignore the friction torque at the pin. The ladder is raised into position by a force applied by a hydraulic piston at C. Point C is a distance 8.00 m from A, and the force F exerted by the piston makes an angle of theta = 40.0 degrees with the ladder.

What magnitude must F have to just lift the ladder off the support bracket at B?

**Imagine a firetruck carrying a horizontal ladder placed on top:

b is close to the center of mass of the ladder but more to the front of the truck

 

[OlderDan]

A ladder of length L = 20.0 m is carried by a fire truck. The ladder has a weight of 2880 N and its center of gravity is at its center. The ladder is pivoted at one end (A) about a pin; you can ignore the friction torque at the pin. The ladder is raised into position by a force applied by a hydraulic piston at C. Point C is a distance 8.00 m from A, and the force F exerted by the piston makes an angle of theta = 40.0 degrees with the ladder.

What magnitude must F have to just lift the ladder off the support bracket at B?

**Imagine a firetruck carrying a horizontal ladder placed on top:

b is close to the center of mass of the ladder but more to the front of the truck

Show us your attempt to solve this problem.

 

[verd]

Hi,

I'm actually having difficulty trying to solve this problem as well... The only difference between her problem and my problem is the weight-- for me, the ladder weighs 2760N. I'm not sure where to start... I know that I have to equate the sum of all the x and y components of F to 0, because the ladder is in equilibrium, but I'm not sure where to go next. I know that I have to find the magnitude of F... So does that mean that when the weight at the end of the ladder is 0, the force is large enough?

Could someone point me in the right direction?

Thank you!

 

[verd]

No help?

 

[Pyrrhus]

I would help, if there was an image, i don't want to work out a incorrect solution.

 

[verd]

Image? Here you go...

http://session.masteringphysics.com/problemAsset/1008055/8/yf_Figure_11_20.jpg

Thanks!

 

[Pyrrhus]

How does the ladder goes up?, let me see if i understand the image, the piston streches pushing the ladder which rotates about point A, and the bar at B is fixed. Right?

 

[Pyrrhus]

Ok B is fixed, the problem practically says it sorry, when the normal force at B is 0 F will have the minimun magnitude to lift the ladder off the B support. Oh yes also take momentum sum with respect to point A, that should give the solution.

 

[verd]

I don't know how I would be able to get the minimum force required to lift the ladder 8m from the pin at a 40 degree angle with the momentum sum... The velocity is going to be 0 until the ladder starts moving...

But what I don't understand how to do is to find the x and y components 8m from the pin... I understand that the Fsin40 and the Fcos40 will be included in both the x's and the y's, but what else is in my force sum? And how to I get to it?


...Thanks...

 

[Pyrrhus]

I don't know how I would be able to get the minimum force required to lift the ladder 8m from the pin at a 40 degree angle with the momentum sum... The velocity is going to be 0 until the ladder starts moving...

But what I don't understand how to do is to find the x and y components 8m from the pin... I understand that the Fsin40 and the Fcos40 will be included in both the x's and the y's, but what else is in my force sum? And how to I get to it?


...Thanks...

8 meters from the pin? the problem does not states that.

It says "What magnitude must F have to just lift the ladder off the support bracket at B?"

Let's look at the momentum sum.

[tex] \sum \tau_{A} = W(\frac{L}{2}) - F \sin \theta (8) - Nd_{BA} [/tex]

But N must be 0, for F to lift it so

[tex] \sum \tau_{A} = W(\frac{L}{2}) - F \sin \theta (8) = 0 [/tex]

 

[verd]

Thanks!

I don't understand how you got [tex]F\sin\theta(8)...[/tex]

However, if I plug in the values with the weight at 2760N, I get [tex]F_{C}=4630.17[/tex], which can't be right...

 

[Pyrrhus]

The distance from C to A is 8 meters, and the force is acting at C. i decomposed the force in 2 components, Fcos(theta) and Fsin(theta). By the way recheck your numeric value.

 

[verd]

Ah. Haha, I was doing this in radians... That'd be my mistake. ...I got it, I got it.

Thanks, by the way. You really helped me out.

 

[Pyrrhus]

No problem, I am glad i helped.