A car on the bridge weighs 2oooN and is 6m away from one of the bridge piers and 4m from the other. How much force does each pier exert to support the car? Pivot point is the weight of the car on the bridge.
combined force is 2 kN
each pier acts as a fulcrum in a class 2 lever system
6 m * 2 kn = 10 m * f1
4 m * 2 kn = 10 m * f2
Well, it seems we have a balancing act on the bridge! Let me calculate the forces at play here. Since the car is acting like a seesaw on the bridge, let's first find the total torque exerted by the car.
The torque (τ) is given by the formula τ = force × distance. For simplicity, let's consider the torque about the pier that is 6m away from the car.
So, the torque about the first pier is τ1 = 2000N × 6m = 12000Nm. Are you still with me? Great!
Now, since the bridge is in equilibrium, the total torque on the other side should be equal and opposite. So, the torque about the second pier is τ2 = -12000Nm. Negative because it's in the opposite direction.
Now, since torque is force times distance, we can rearrange the equation to find the force exerted by each pier. Let's call the force exerted by the first pier F1 and by the second pier F2.
So, for the first pier, we have τ1 = F1 × 6m. Solving for F1, we find that F1 = 12000Nm / 6m = 2000N.
And for the second pier, we have τ2 = F2 × 4m. Solving for F2, we find that F2 = -12000Nm / 4m = -3000N.
Now, I must confess, negative forces sound a bit mysterious, but don't worry, it just means that the second pier is exerting force in the opposite direction to balance everything out.
So, in conclusion, the first pier is exerting a force of 2000N to support the car, while the second pier is exerting a force of -3000N. Just a little balancing act keeping everything in check!
To find out how much force each pier exerts to support the car on the bridge, we can use the principle of moments. The total moment on a body in equilibrium is equal to zero.
In this case, the weight of the car on the bridge is located at its center, so the pivot point is the weight of the car on the bridge. Let's assume that the force exerted by the left pier is F1, and the force exerted by the right pier is F2.
The moment from the left pier can be calculated as the force (F1) multiplied by the distance to the pivot point (6m), and the moment from the right pier is calculated as the force (F2) multiplied by the distance to the pivot point (4m).
Since the bridge is in equilibrium, the total moment from both piers must be equal to zero. Mathematically, this can be expressed as:
(F1 * 6m) + (F2 * 4m) = 0
Now, let's solve for F1 and F2:
6F1 = -4F2
Divide both sides by 6:
F1 = -4F2/6
Simplifying:
F1 = -2F2/3
Alternatively, we can also use the principle of moments at the pivot point:
The total clockwise moment from the left pier will be equal to the total anticlockwise moment from the right pier, as both moments will balance each other out:
(F1 * 6m) = (F2 * 4m)
6F1 = 4F2
Simplifying:
F1 = 2F2/3
Now that we have the relationship between F1 and F2, we can solve for their values.
Let's assume F2 as some arbitrary value, such as 1000N:
F1 = 2(1000N)/3
F1 = 2000N/3
F1 ≈ 666.67N
So, each pier exerts a force of approximately 666.67N and -1000N (as it is in the opposite direction) to support the car on the bridge.
To determine the force exerted by each pier to support the car on the bridge, we can use the principle of moments. The principle of moments states that the sum of the clockwise moments about any point is equal to the sum of the anticlockwise moments about the same point.
In this case, we will take the pivot point as the weight of the car on the bridge. Since the car weighs 2000 N, the pivot point will also exert a downward force of 2000 N.
Let's calculate the moments about the pivot point for each pier:
For the pier that is 6 m away from the pivot point:
Moment1 = Force1 × Distance1
Moment1 = Force1 × 6 m
For the pier that is 4 m away from the pivot point:
Moment2 = Force2 × Distance2
Moment2 = Force2 × 4 m
According to the principle of moments, the sum of the moments about the pivot point equals zero since the bridge is in equilibrium:
Sum of moments = Moment1 + Moment2
= Force1 × 6 m + Force2 × 4 m
Since the force exerted by each pier is vertical (upwards) in this case, we can equate the sum of the moments to zero:
Force1 × 6 m + Force2 × 4 m = 0
Now, we know that Force1 + Force2 = 2000 N (the total weight of the car on the bridge).
We can substitute this in the equation:
Force1 × 6 m + (2000 - Force1) × 4 m = 0
Now, we can solve this equation to find the values of Force1 and Force2.
Simplifying the equation:
6Force1 + 4(2000 - Force1) = 0
6Force1 + 8000 - 4Force1 = 0
2Force1 + 8000 = 0
2Force1 = -8000
Force1 = -4000 N
We get a negative value for Force1, which indicates that the force is acting in the opposite direction (downwards) compared to our assumed direction (upwards). To obtain the magnitude, we take the absolute value:
|Force1| = |-4000 N| = 4000 N
Now, we can calculate the value of Force2:
Force2 = 2000 N - Force1
Force2 = 2000 N - 4000 N
Force2 = -2000 N
Similarly, the magnitude of Force2 is:
|Force2| = |-2000 N| = 2000 N
Hence, each pier exerts a force of 4000 N and 2000 N, respectively, to support the car on the bridge.