A bridge 100m long weighs 500kN. A lorry weighing 100kN is 25m from one end of it. Find the force exerted at this support?
Well, if you insist on asking the tough questions, I suppose I'll have to crack a few jokes to lighten the mood. Now, let's get to the bottom of this bridge dilemma, shall we?
To find the force exerted at the support, we need to consider the weight distribution on the bridge. The lorry weighs 100kN and is positioned 25m from one end, which means it has a moment arm of 25m.
Now, let's put on our mathematical hats and do some calculations. We know that the total weight of the bridge is 500kN, and the lorry's weight is 100kN. If we assume the weight is evenly distributed along the length of the bridge, we can find the force at the support.
Since the lorry is 25m from the support, that means the remaining length of the bridge is 75m. Therefore, the weight of the remaining part of the bridge is (500kN - 100kN) = 400kN.
To find the force at the support, we multiply the weight of the remaining part of the bridge by its moment arm. Therefore, the force exerted at the support is (400kN * 75m) / 100m = 300kN.
So, after all that math, the force at the support is 300kN. I hope this answered your question, and remember, when it comes to bridges, it's all about finding the right balance... just like in life!
To find the force exerted at the support, we need to consider the moments acting on the bridge. A moment is the turning effect of a force.
We can start by calculating the total moment on the bridge. The total moment at any point is given by the product of the force and the perpendicular distance from the point to the line of action of the force.
In this case, the total moment is zero because the bridge is in equilibrium and not rotating. Therefore, the sum of the clockwise moments must be equal to the sum of the anticlockwise moments.
Let's denote the force exerted at the support as F1. The distance of the lorry from the support is 25m. We can find the moment of the lorry by multiplying its weight by this distance:
Moment of the lorry = 100kN * 25m
As the bridge is in equilibrium, this moment should cancel out the anticlockwise moments acting on the bridge. The bridge's weight is given as 500kN, and its length is 100m. The weight of the bridge can be considered to act at its center, which is halfway between the supports. Therefore, the perpendicular distance between the support and the line of action of the bridge's weight is 50m.
Moment of the bridge = 500kN * 50m
Now, considering equilibrium, we can set up the equation:
Total anticlockwise moment = Total clockwise moment
Moment of the bridge + Moment of the lorry = F1 * 100m
(500kN * 50m) + (100kN * 25m) = F1 * 100m
Simplifying:
25000 + 2500 = F1 * 100
27500 = F1 * 100
To find F1, we can divide both sides by 100:
F1 = 27500 / 100
F1 = 275 kN
Therefore, the force exerted at the support is 275 kN.
To find the force exerted at the support, we can use the principles of static equilibrium. In static equilibrium, the sum of the forces and the sum of the torques acting on an object are zero.
Let's first find the force exerted by the lorry. We know that the weight of the lorry is 100 kN, and it is located 25 m from one end of the bridge. In order to maintain static equilibrium, the force exerted by the lorry must be countered by an equal and opposite force exerted by the support at this end of the bridge.
Let's assume that the force exerted by the support is F (in kN). Since the lorry is 25 m from the support, the torque exerted by the lorry is equal to the force exerted by the lorry multiplied by the distance from the support.
Torque = Force x Distance
The torque exerted by the lorry is given by:
10 kN (100 kN / 10) x 25 m = 250 kN·m.
To maintain static equilibrium, the torque exerted by the support must be equal and opposite to the torque exerted by the lorry. Since the support is located at the same end as the lorry, the torque exerted by the support would be in the clockwise direction, opposite to the lorry's torque.
Therefore, the torque exerted by the support is -250 kN·m.
Now, let's consider the sum of the torques acting on the bridge. Since there are no other forces acting on the bridge, the sum of the torques must be zero.
The sum of the torques = Torque exerted by the lorry + Torque exerted by the support
0 = 250 kN·m + (-250 kN·m)
This implies that the torque exerted by the support is -250 kN·m.
Finally, the force exerted by the support can be found by dividing the torque exerted by the support by the distance from the support.
Force = Torque / Distance
Force = -250 kN·m / 100 m = -2.5 kN
Therefore, the force exerted at the support is -2.5 kN. The negative sign indicates that the force is directed in the opposite direction as assumed earlier.