Physics.Statics triangle of force

bluestreak

Hi just wondering if i could get some help on this question.

A ladder leans against a perfectly smooth vertical wall at 30 degrees to the horizontal, a load of 800 N is placed three quarters of the way up the ladder. if the ladder rests on a rough horizontal surface which prevents slipping, find the magnitude and the direction of the reaction between the ladder and the ground..?????

Lu Tze

Ill give you a start at least (been a while since i have done these)

Assuming the ladder stays in place there is an opposing reaction from the ground and wall. The wall reaction is horizontal only as it is perfectly smooth.

The ground reaction is horizontal and vertical.

You should be able to break up the 800N force into vertical and horizontal components and solve from that.

Edit: not sure whether moments come into play here given that there is a distance up the ladder specified - might want to hang on until somebody who has done these in the last ten years comes on!

Morbert

bluestreak wrote: »

Hi just wondering if i could get some help on this question.

A ladder leans against a perfectly smooth vertical wall at 30 degrees to the horizontal, a load of 800 N is placed three quarters of the way up the ladder. if the ladder rests on a rough horizontal surface which prevents slipping, find the magnitude and the direction of the reaction between the ladder and the ground..?????

From what I am inferring, it is asking you to determine the normal force.
http://en.wikipedia.org/wiki/Normal_force

To get you started:

Divide the force into components perpendicular and parallel to the ladder.

The normal force will be equal in magnitude and opposite in direction to the perpendicular component.

bluestreak

thanks have found the reactions, but where do i go from here????

Morbert

bluestreak wrote: »

thanks have found the reactions, but where do i go from here????

Once you have the reaction, that should be it. You've solved the problem and don't need to do anything more.

Have you divided the 800N force into perpendicular and parallel components? The reaction force will be equal in magnitude and opposite in direction to the perpendicular component.

FISMA

This sounds like the standard homework question for the Sum of the Torques (SoT's). You are learning rotational motion and torques, correct?

Step 1: draw a free body diagram of the Forces ON the ladder.
Step 2: do the Sum of the Forces (SoF's) in the x direction
Step 3: do the Sum of the Forces (SoF's) in the y direction
Step 4: do the SoT's for the ladder, about the contact pt with the ground.
Step 5: solve

Did they give you the mass of the ladder? Probably okay to neglect it given your 500kg mass.

A bit of help for the FBD

Since there's no friction between ladder and wall, the house pushes horizontally on the ladder with a Normal Force.

The weight of the ladder acts down from the CoM.

The weight due to the mass acts down from its pt of application.

There's a normal force at the bottom of the ladder, since it does not sink into the ground. Also, there's a frictional force that acts towards the wall.

If you write back and complete the steps as shown (and show your work), I'll help you finish it up. There's actually an easier way to resolve the Forces in the SoT's.

Slan

Morbert

Jaysus, completely misread the question. Thought it was asking for the reaction between the ladder and the load.