Simple ladder question (moments)

[bonbon22]

Homework Statement

A ladder of mass 20 kg and 3.0 m in length is propped up against a wall at an angle of 72o.

Calculate the support force on the wall.

Which data item is irrelevant?

g = 9.81 N kg-1.

Relevant Equations

f*d= moment

I know the answer and how to go about it the normal way. Could i possibly use the nose to tail method of finding the S value. Although it does not say in the question there are three forces on ladder support force , weight, support form the wall. So could i make a triangle then use sine rule cause i think that is so much easier. I used method and got s/sin(18) = 20 g/sin(72) and i got s as 63 the answer is 31

 

[haruspex]

there are three forces on ladder support force , weight, support form the wall.

What exactly do you mean by the first of those three? What direction is it in?

 

[bonbon22]

What exactly do you mean by the first of those three? What direction is it in?

the support force from the ground against the ladder. imagine the ladder leaning against the wall. The direction of that one since its in equilibrium would be 72 degrees to the horizontal cause mg acts straight down. support force from the wall is to the left.

 

[haruspex]

The direction of that one since its in equilibrium would be 72 degrees to the horizontal

I see no reason it should be at that angle.
Why doesn't it fall down?

 

[bonbon22]

I see no reason it should be at that angle.
Why doesn't it fall down?

why would it fall down? the ladder makes an angle of 72 as the question said so I am assuming its 72 why is that not the case?
http://www.antonine-education.co.uk/Pages/Physics_2/Mechanics/MEC_03/Mechanics_3.htm
if you scroll down you can see the question and the normal way of getting the answeer.

 

[Chestermiller]

Is there friction at the wall?

 

[bonbon22]

Is there friction at the wall?

doesn't say. the solution from the link if you can't click it.Weight = 20.0 kg × 9.81 N kg-1 = 196.2 N

S = (196.2 N × cos 72) ÷ (2 × sin 72) = 31.9 N



The length is irrelevant.

 

[Chestermiller]

So are you saying that you don't know how this solution came about?

 

[bonbon22]

So are you saying that you don't know how this solution came about?

No mate. I understand the normal way of going about it, i was asking if you can use the " nose to tail method" of finding the S value or is that not a method i can use? i find that way easier so i was asking. I got an answer which is twice as big as the actual answer. Is the method wrong or the way i did it incorrect?the way i did it:

You can see an upside down triangle, this is just an image to make it easier to understand how i got my answer. Question does not mention frictional force. The upside down triangle has an angle on the bottom side theta 72, so using the Z angle at point B it also has 72, so i can use soh cah toa to get the top of the triangle the S force acting against the ladder from the wall. I get an answer which is twice as big what did i do wrong?

 

[Chestermiller]

Did you get your result by taking moments about point A? Otherwise, please show us your equation.

 

[bonbon22]

Did you get your result by taking moments about point A? Otherwise, please show us your equation.

i made a triangle ... if you read my previous reply friendo

using soh cah toa1) Mg / S = tan(72)

Mg/ tan (72) = S
M = 20 g = 9.8
S= 63
answer is 31

 

[Chestermiller]

Your equation assumes (incorrectly) that the force at A is directed parallel to the ladder. Now that you've tried the "guessing" method, let's see your force- and moment balance equations.

 

[haruspex]

No mate. I understand the normal way of going about it, i was asking if you can use the " nose to tail method" of finding the S value or is that not a method i can use? i find that way easier so i was asking. I got an answer which is twice as big as the actual answer. Is the method wrong or the way i did it incorrect?the way i did it:View attachment 243016
You can see an upside down triangle, this is just an image to make it easier to understand how i got my answer. Question does not mention frictional force. The upside down triangle has an angle on the bottom side theta 72, so using the Z angle at point B it also has 72, so i can use soh cah toa to get the top of the triangle the S force acting against the ladder from the wall. I get an answer which is twice as big what did i do wrong?

You say you assumed the resultant of the two forces from the ground is along the line of the ladder. If that is the case, what do you notice if you take moments about either B or the middle of the ladder?

 

[bonbon22]

You say you assumed the resultant of the two forces from the ground is along the line of the ladder. If that is the case, what do you notice if you take moments about either B or the middle of the ladder?

i get it now cheers m8

 

[haruspex]

i get it now cheers m8

There's a handy rule: if three forces on a body are in equilibrium then their lines of action must pass through a common point.