What are the normal forces at points A and B in a moment and couple problem?

[yoda2026]

ive been trying to figure our this mechanics question and I am rather stuck

but unfortunately i haven't had much experience with moments

find the normal forces applied to the installed stud at points A and B
l-----360mm--l--85-l
A.====
P.=========== /==/
====.B

length from P to A = 360MM
length of A to B = 85mm
thickness of timber between A and B is 45mm (or vertical distance between A and B)

for applied at point P is 150N

ignore friction


so far i understand that the force applied to the centre of the stud
is 360/1000 * 150N = 54NM

then you have a couple of 85/1000 * 150N = 12.75nm

is this the correct way of attempting this question
if so how does the couple and the for applied to the lever interact to give me the force on points A and B of the stud?

 

[tiny-tim]

welcome to pf!

hi yoda2026! welcome to pf!

if the question is what i think it is , then you find the force at A by taking moments about B …

the moment about B of the force at P must be equal and opposite to the the moment about B of the force at A

 

[yoda2026]

so when you say the moment of about B
do you mean the moment of PB or the moment AB

 

[tiny-tim]

so when you say the moment of about B
do you mean the moment of PB or the moment AB

i'm sorry, i don't understand …

the moment of a force F about a point B is F times the perpendicular distance from B to the line of the force

 

[yoda2026]

find the normal forces applied to the installed stud at points A and B
l-----360mm--l--85-l
.....A.====
P.========== /==/
.....====.B

sorry my diagram lost its formatting
i am trying to calculate for the force on points A and B

now the way i have been looking at it is that the force is 150N is applied to point P
so the moment between PA is 360mm x 150N = 54nm

this is where i get abit lost because points A and B form a couple producing 12.5NM
is the force applied to point B calculated from PB or is it PA-AB

or have i just got this totally backward

 

[tiny-tim]

hi yoda2026!

(just got up :zzz: …)

… now the way i have been looking at it is that the force is 150N is applied to point P
so the moment between PA is 360mm x 150N = 54nm

your language is completely wrong, which i think is why you're misunderstanding how moments work

there is no "moment between PA" …

there is a moment of the force at P about the point A …

in this case, the force is vertical, so the moment is the force times the horiztonal distance between P and A

ok, now you need the moments of all the other forces about A

clearly, the force at A has zero moment

and the moment of the force at B about the point A is … ? ​

 

[yoda2026]

im guessing it is 66.25nm

because the force experienced in the couple is equal in both directions
now to calculate the normal force i am totally stumped

because all these cals are in NM(which is a rotational force) and please correct me if I am wrong but a normal force is a linear force and would be measured in Newtons

 

[yoda2026]

[PLAIN]http://img851.imageshack.us/img851/1826/photoc.jpg

here is a actual copy of the diagram
i hope this makes it clearer

 

[tiny-tim]

(thanks for putting the diagram sideways … i hate it when people post diagrams that are too wide! )

im guessing it is 66.25nm

how did you get that?

because all these cals are in NM(which is a rotational force) and please correct me if I am wrong but a normal force is a linear force and would be measured in Newtons

yes

because the force experienced in the couple is equal in both directions
now to calculate the normal force i am totally stumped

there isn't only a couple

at both A and B, there is a vertical force (the two normal forces which you are asked to find)

they are not equal: they form a couple plus a resultant linear force of 150 N (to balance the force on the handle)

let's say that the vertical force at B is V Newtons …

how much is the moment of that force about A, and so what is V ?