[ani-q]

Questio one

A uniform rod AB has lenght 4m and weight 150 N.
Rod rests in equilibrium in a horizontal position, smoothly supported at points C & D where AC=1m and AD= 2.5m as shown in the diagram above.
A particle of weight WN is attatched to the rod at point E where AE=x metres.
Rod remains in equilibrium and the magnitude of the reaction at C is now equal to the mgnitude of the reaction at D

a) show that W=150/7-4x - dont know how to do either parts

b) Hence deduce the range of possible values of x


Qurestion 2

A plank AB of mass 12kg and lenght 3m is in equilibrium in a horizontal position resting on supports C&D where AC=0.7m and DB=1.1m
A boy of mass 32kg stands on the plank at point E.
The plank is about to tilt about D
By modelling the plank as a uniform rod and the boy as a particle, calculate the distance AE??



last question

A uniform rod AB has lenght 5m and weight 20 N
The rod is resting on supports at point C and D where AC =2m and BD=1m
a) find the magnitudes of the reactions at C and D

A particle of weight 12 N is placed on the rod at point A
b) show that this causes the rod to tilt about C

A second particle of weight 12N is placed on the rod at E to hold it in equilibrium
c) how far must E be from A?

pt a i did but pt b i got stuck on and therefore couldnt do part c


thanks before hand

[I'm JoeKing]

Wait for more people to answer; I'm quite the failure at Maths, and need revision on it as much as anyone does lmao. Here are my, probably awful, attempts.

Q1 I am completely clueless. Good start, then.

Q2:
About A: 1.5*12g=(3-x)32g
so 18g=96g-32xg
so78g=32xg
cancel out g
78=32x
x=78/32=2.4M (2s.f)
so length AE=2.4M(2s.f)

^But that all seems too easy. And Maths isn't easy. So I'm probably wrong.


Q3a.
Moments about C:
0.5*20=2*Rd
10N=2*Rd
5N=Rd
Reaction at D is 5N

Moments about D:
1.5*20N=2*Rc
30N=2*Rc
15N=Rc
Reaction at C is 15N
(and Reaction at D is 5N)

3b.
Moments about C: (2*12N)+(2*Rd)-(0.5*20N)
Moments about C: 24N+10N-10N
Moments about C: 24N (anticlockwise)
Therefore, the rod tilts about C with an anticlockwise moment of 24Nm.

3c.
Moments about C: (x * 12N) must equal the 24Nm anticlockwise moment
x*12N=24Nm
x=24Nm/12N
x=2M
So E is 2m from C, which is 2m from A
so AE=2m+2m
Length A3 = 4m.


Yeahhhh, don't go writing all that down anytime soon, because it's all very probably embarrassing and wrong, but that's my attempt nonetheless :')

[ani-q]

3b, was right aha but thanks anyway