A hiker, who weighs 976 N, is strolling through the woods and crosses a small horizontal bridge. The bridge is uniform, weighs 4350 N, and rests on two concrete supports, one on each end. He stops 1/3 of the way along the bridge. What is the magnitude of the force that a concrete support exerts on the bridge (a) at the near end and (b) at the far end?
To find the magnitude of the force that a concrete support exerts on the bridge, we need to analyze the forces acting on the bridge and apply the principles of equilibrium.
Let's start with the forces acting on the bridge:
1. The weight of the hiker (976 N) acts vertically downward.
2. The weight of the bridge itself (4350 N) also acts vertically downward.
3. The reaction forces from the concrete supports act upwards from each end of the bridge.
Now, let's break it down into two scenarios: one for each end of the bridge.
(a) At the near end:
Since the hiker is 1/3 of the way along the bridge, only 1/3 of the hiker's weight is acting on the near end. Therefore, the hiker's weight that acts on the near end is (1/3) * 976 N = 325.33 N.
To find the magnitude of the force that the concrete support exerts on the near end, we need to balance the forces vertically. At equilibrium, the sum of the vertical forces is zero.
So, the equation will be:
Reaction force at the near end - 325.33 N (hiker's weight) - 4350 N (bridge's weight) = 0
Simplifying the equation:
Reaction force at the near end = 325.33 N + 4350 N
Therefore, the magnitude of the force that the concrete support exerts on the near end is 325.33 N + 4350 N = 4675.33 N.
(b) At the far end:
Since the hiker is 1/3 of the way along the bridge, 2/3 of the hiker's weight is acting on the far end. Therefore, the hiker's weight that acts on the far end is (2/3) * 976 N = 650.67 N.
Applying the same equilibrium principle:
Reaction force at the far end - 650.67 N (hiker's weight) - 4350 N (bridge's weight) = 0
Simplifying the equation:
Reaction force at the far end = 650.67 N + 4350 N
Hence, the magnitude of the force that the concrete support exerts on the far end is 650.67 N + 4350 N = 5000.67 N.
Let's start by calculating the weight of the hiker. The weight (W) is given as 976 N.
Now let's calculate the weight of the bridge. The weight of the bridge (WB) is given as 4350 N.
Since the hiker stops 1/3 of the way along the bridge, the distance from the near end to the hiker is 1/3 of the total length of the bridge.
We can use the concept of torque to determine the forces exerted on the bridge by the supports. Torque is the product of force and perpendicular distance from the point of rotation.
(a) To find the force at the near end, we need to consider the torques acting about the near end of the bridge.
The torque exerted by the hiker's weight is given by T1 = W * d1, where W is the hiker's weight and d1 is the distance from the near end to the hiker.
The torque exerted by the bridge's weight is given by T2 = WB * d2, where WB is the weight of the bridge and d2 is the distance from the near end to the far end of the bridge.
Since the bridge is uniform, its center of mass is at the middle. Therefore, d1 = d2/2.
The total torque about the near end is T_total = T1 + T2.
(b) Similarly, to find the force at the far end, we need to consider the torques acting about the far end of the bridge.
The torque exerted by the hiker's weight is still given by T1 = W * d1.
The torque exerted by the bridge's weight is given by T3 = WB * d3, where d3 is the distance from the far end to the hiker.
The total torque about the far end is T_total = T1 + T3.
Now we can plug in the given values and calculate the forces.
Let's assume the length of the bridge is L.
(a) The distance d1 = (1/3)L, and since d1 = d2/2, we have d2 = 2 * (1/3)L = (2/3)L.
T1 = W * d1 = 976 N * (1/3)L.
T2 = WB * d2 = 4350 N * (2/3)L.
T_total = T1 + T2.
(b) The distance d3 = (2/3)L - (1/3)L = (1/3)L.
T3 = WB * d3 = 4350 N * (1/3)L.
T_total = T1 + T3.
To find the magnitude of force at each end, divide the respective torques by the distance from the support to the end of the bridge.
(a) The force at the near end is F_near = T_total / ((1/3)L).
(b) The force at the far end is F_far = T_total / ((2/3)L).
Simplifying the equations gives:
(a) F_near = (976 N * (1/3)L + 4350 N * (2/3)L) / ((1/3)L).
(b) F_far = (976 N * (1/3)L + 4350 N * (1/3)L) / ((2/3)L).
Now you can simplify the above equations further and calculate the magnitude of the forces at each end.