A hiker, who weighs 915 N, is strolling through the woods and crosses a small horizontal bridge. The bridge is uniform, weighs 4185 N, and rests on two concrete supports, one at each end. He stops one-fifth of the way along the bridge. What is the magnitude of the force that a concrete support exerts on the bridge at each end?
To find the magnitude of the force that a concrete support exerts on the bridge at each end, we first need to understand the concept of equilibrium. In equilibrium, the sum of all forces acting on an object must be equal to zero.
In this scenario, the hiker exerts a downward force due to his weight, and the bridge exerts an upward force to counterbalance it. Additionally, the concrete supports exert vertical forces to support the bridge's weight.
Since the hiker is one-fifth of the way along the bridge, let's call the length of the bridge "L". Therefore, the distance from each support to the hiker is (1/5)L.
To solve this problem, we will set up the equation for vertical equilibrium. The sum of all vertical forces must be equal to zero:
ΣF_vert = 0
Let's consider the forces involved:
1. The hiker's weight is 915 N, and it acts downward.
2. The bridge's weight is 4185 N, and it acts downward.
3. The support forces from the concrete supports act upward, and we'll call these F_support.
Summing up the forces in the vertical direction:
915 N + 4185 N + 2F_support = 0
Now, let's rearrange the equation to solve for F_support:
2F_support = - (915 N + 4185 N)
2F_support = - 5100 N
F_support = - 2550 N
The negative sign indicates that the concrete supports exert an upward force to balance the downward forces of the hiker and the bridge.
However, the magnitude of the force cannot be negative, so we take the positive value:
Magnitude of F_support = |-2550 N| = 2550 N
Therefore, the magnitude of the force that a concrete support exerts on the bridge at each end is 2550 N.