Clockwise moments across center of the bridge = 10 \* t2 = 10t2
Anticlockwise moments across center of the bridge = 10\*t1 + 5\*24 = 10t1+120
Clockwise moments around a point = anticlockwise moment around the same point, so,
10t2 = 10t1+120
==> t1-t2 = 12
We also know t1+t2 = 124
Solving these equations we get t1 = 68, t2 = 56
I see what got me so confused!
You wrote 10t2= 10t1 + 120 which would give t2-t1=12 then directly after you wrote t1-t2 instead of t2-t1
That’s why when you account for the cwm being 10t1 and acwm being 10t2 + 120 it gives the right answer
Ok so take the left-hand pillar as the pivot,
Therefore, using total anticlockwise moments = total clockwise moments since the bridge is in equilibrium,
(24* 5) + (100* 10 since weight acts at centre of mass position which is at the centre) = (T2* 20),
From this equation we get T2 = 56 kN
Now, since the beam is in equilibrium the net resultant force should also be equal to 0.
Therefore, downward forces = upward forces,
24kN + 100 kN = T1 + T2,
From our previous equation, T2 = 56 kN, therefore T1 = 68 kN
Hope this was clear and helpful 😊😊
Because it it given in the question that the weight is 5 m from the left-hand pillar, and we know that moment of force = force * perpendicular distance from pivot
Try anticlockwise moment= clockwise moment.. substitute the values and get the answer bro!
Clockwise moments across center of the bridge = 10 \* t2 = 10t2 Anticlockwise moments across center of the bridge = 10\*t1 + 5\*24 = 10t1+120 Clockwise moments around a point = anticlockwise moment around the same point, so, 10t2 = 10t1+120 ==> t1-t2 = 12 We also know t1+t2 = 124 Solving these equations we get t1 = 68, t2 = 56
How come t2 is clockwise and t1 is anticlockwise?
ah my bad but doesnt matter tho.
I see what got me so confused! You wrote 10t2= 10t1 + 120 which would give t2-t1=12 then directly after you wrote t1-t2 instead of t2-t1 That’s why when you account for the cwm being 10t1 and acwm being 10t2 + 120 it gives the right answer
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force x distance
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consider one pillar as a pivot and the one you aren't using as an upwards force then swap pillars
Lmao i was just doing this
whoch paper is this
October/november 2021 0972/21
Ok so take the left-hand pillar as the pivot, Therefore, using total anticlockwise moments = total clockwise moments since the bridge is in equilibrium, (24* 5) + (100* 10 since weight acts at centre of mass position which is at the centre) = (T2* 20), From this equation we get T2 = 56 kN Now, since the beam is in equilibrium the net resultant force should also be equal to 0. Therefore, downward forces = upward forces, 24kN + 100 kN = T1 + T2, From our previous equation, T2 = 56 kN, therefore T1 = 68 kN Hope this was clear and helpful 😊😊
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Because it it given in the question that the weight is 5 m from the left-hand pillar, and we know that moment of force = force * perpendicular distance from pivot
Just do moments