# M2 statics of rigid bodies

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#1
I hate this topic, can do the ladders questions but the other ones bug me 3: (no picture )
The figure shows a pole of mass m and length l displaying a light banner. The pole is modelled as a uniform rod AB, freely hinged to a vertical wall at point A. It is held in horizontal position by a light wire inclined at angle & to the horizontal. One end of this wire is attached to the end B of the rod and the other end is attached to the wall at point C which is vertically above A and such that tan & = 1/3

a) Show that the tension in the wire is 0.5mg x sqrt10

Ok, I got this by doing moments about A. However I first tried resolving vertically and I got mg x sprt10. Why didnt resolving vertically work? Am i missing a force with a vertical component? I have the tension and mass but thats it.

b) Find the magnitude of the force exerted by the wall on the rod at A

Is this the horizontal force acting along the pole at A? I tried resloving horizontally but that didnt work. Am I missing any forces???
0
15 years ago
#2
(Original post by Womble548)
I hate this topic, can do the ladders questions but the other ones bug me 3: (no picture )
The figure shows a pole of mass m and length l displaying a light banner. The pole is modelled as a uniform rod AB, freely hinged to a vertical wall at point A. It is held in horizontal position by a light wire inclined at angle & to the horizontal. One end of this wire is attached to the end B of the rod and the other end is attached to the wall at point C which is vertically above A and such that tan & = 1/3

a) Show that the tension in the wire is 0.5mg x sqrt10

Ok, I got this by doing moments about A. However I first tried resolving vertically and I got mg x sprt10. Why didnt resolving vertically work? Am i missing a force with a vertical component? I have the tension and mass but thats it.

b) Find the magnitude of the force exerted by the wall on the rod at A

Is this the horizontal force acting along the pole at A? I tried resloving horizontally but that didnt work. Am I missing any forces???

is this in the orange m2 book? if so what page?
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#3
Its in the 'revise for' book. Exam q3 at end
0
15 years ago
#4
dont have it damn..sorry i cant visualize the pic
0
15 years ago
#5

Hope thats an acurate enough diagram...

Anyway... way to do it would be:

l.mg/2 = l.T√10

so cancel l and solve for T.. and you get:

T = 0.5mg * √10

-------------------------------
b) Resolve Horizontally for this (I think) - the number of marks for the question often gives you the hint.

so Normal Reaction at A (R)= T.cos(α)

So,

R = 0.5mg*√10*3/√10

R = 1.5mg

Note, I'm not 100% sure about part B.. whats the answer in the book?

MdSalih
0
15 years ago
#6
lol I am doing revising for M2 by looking at people's posts on TSR :P
I suck at M2 by the way, so dun trust my words, thank you~

T = 0.5mg*sqrt10

At hinge A there are two forces, one vertical and one horizontal (I think)

Resolve vertical
R = mg + 0.5mg*sin(α)*sqrt10
= 15.mg

Resolving horizontal
F = 0.5mgXsqrt10X cos(α)
= 1.5mg

I got R = F = 1.5mg at the hinge..

and very uncertain on the following part as to whether correct or not and whether it is actually needed:

magnitude (P) of force at hinge:

P = sqrt ( (1.5mg)^2 + (1.5mg)^2)
= sqrt (432.18m^2)
= 20.78893937 X m
so hinge force has magnitude = 20.8m at direction 45deg to horizontal?
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#7
The reaction at the wall is 0.5mg * √10

Its a 7mark question to find the reaction so I guess there is more work than resolving horizontally I did that and got the same as you.

Ill just have to hope that a ladders question comes up instead
0
15 years ago
#8
(a)
Vertically: Tsint + Y = mg
Tcost = X

(Tsint)(l) = (mg)(.5l)
Tsint = 0.5mg

T = mg/2sint
T = mg/2sin(arctan1/3)
T = 0.5mgrt10

(b) X = Tcost = 3mg/2
Y = mg-Tsint = 0.5mg

Force = mgrt[1/4 + 9/4]^.5
= 0.5mgrt10
0
#9
Ahhh, got ya! There is more than just a horizontal force at A.

So that explains why resolving vertically for Q1 didnt work!
0
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