# lads can only do this m2 question

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#1
it's 5a first question in heinemann book

a uniform beam AB of mass 20 kg and length 2m is attached to a vertical wall by means of a smooth hinge at A. The beam is maintained in the horizontal psition by means of a light inextensible string, one end of which is attached to the beam at B and the other end is attached to the wall at a point in the wall 2m vertically above A

a) calc T
0
15 years ago
#2
I'll go... T = 139N
Cud be horribly wrong, ill check my genitalia in a moment err!
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15 years ago
#3
(Original post by YorkshirePhoeni)
I'll go... T = 139N
Cud be horribly wrong, ill check my genitalia in a moment err!
i get t'same
0
15 years ago
#4
ditto;

if you take moments about A, you have:-
20g = 2Tsin45
T= 10g / sin45 = 138.6N
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#5
wt if tension in string was 400 n n there was a mass at b
wtn is da mass?
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15 years ago
#6
(Original post by mabs)
wt if tension in string was 400 n n there was a mass at b
wtn is da mass?
Take moments about A again and call the mass at b 'M', you already know the tension so it should come out using the moments equation.
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#7
everyone knows the method but wots the ansa darlin???
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15 years ago
#8
(Original post by mabs)
wt if tension in string was 400 n n there was a mass at b
wtn is da mass?
As Mysticmin said, using the same format; Moments about A

20g + 2M = 400 sin 45

2M = 400 sin 45 - 20g

M = (400 sin 45 - 20g)/2

M = 43.4N

Therefore Mass M = 43.4/9.8 = 4.43 kg (Do not forget this step)
0
15 years ago
#9
(Original post by YorkshirePhoeni)
As Mysticmin said, using the same format; Moments about A

20g + 2M = 400 sin 45

2M = 400 sin 45 - 20g

M = (400 sin 45 - 20g)/2

M = 43.4N

Therefore Mass M = 43.4/9.8 = 4.43 kg (Do not forget this step)
I got a different answer by taking moments, ive probably gone wrong somewhere, if so , feel free to point out and **** off!...

moments from A... letting mass at B =B

20gm + 2mBg = 400sin45x2m

remove 2m from all...

10g + Bg = 400sin45
98 + 9.8B = 282
9.8B = 184
B = 18.86 kg
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#10
Taking moments
===========

2m.Tcos45 - 2m.Mg = 1m.20g
2.400.1/√2 - 2Mg = 20g
Mg = 200√2 - 10g
M = 200√2/g - 10
M = 28.86 - 10
M = 18.86
M = 18.7 kg (3 sf)
========
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#11
like i sed only men can do m2
0
15 years ago
#12
(Original post by mabs)
===========

M = 18.86
M = 18.7 kg (3 sf)
========
I liked the answer before the strange sig figs...
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#13
kk georgey

we having fun

girls arre wallowin in self pity
0
15 years ago
#14
(Original post by mabs)
like i sed only men can do m2
Hove you done it yet? what did you get?
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15 years ago
#15
(Original post by mabs)
like i sed only men can do m2
You've got to be bloody joking. I'm a girl and I have never got below 65/75 in M2.
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15 years ago
#16
(Original post by Mysticmin)
You've got to be bloody joking. I'm a girl and I have never got below 65/75 in M2.
Do you get the joke or not? lol
0
15 years ago
#17
(Original post by George-W-Duck)
I got a different answer by taking moments, ive probably gone wrong somewhere, if so , feel free to point out and **** off!...

moments from A... letting mass at B =B

20gm + 2mBg = 400sin45x2m

remove 2m from all...

10g + Bg = 400sin45
98 + 9.8B = 282
9.8B = 184
B = 18.86 kg
Oh aye forgot the 2m x the 400 sin 45, else i get the same 18.86
This is the start of my m2 revision i DO apologise!
0
15 years ago
#18
(Original post by glance)
Do you get the joke or not? lol
Nah, I only saw that post, explaination?
0
15 years ago
#19
(Original post by glance)
Do you get the joke or not? lol
I dont' get the joke...explain
0
15 years ago
#20
Maybe there wasn't one. I didn't think it was very funny...anyway, forget that.
0
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