A wooden bridge crossing a canyon consists of a plank with length density λ = 2 kg/m suspended at h = 10.0 m below a tree branch by two ropes of length L = 2h and with a maximum rated tension of 2000 N, which are attached to the ends of the plank, as shown in the figure. A hiker steps onto the bridge from the left side, causing the bridge to tip to an angle of 25° with respect to the horizontal. What is the mass of the hiker? (well explained please)
I need help
To find the mass of the hiker, we need to consider the forces acting on the system and analyze the given information.
First, let's identify the forces acting on the system:
1. Weight of the plank: The plank is suspended vertically, so it experiences a downward force due to gravity. The weight of an object is given by the formula W = m * g, where m is the mass and g is the acceleration due to gravity. In this case, the plank's weight is W_plank = λ * L.
2. Tension in the ropes: The ropes are attached to the ends of the plank and are under tension due to the weight of the plank and the hiker. The maximum rated tension for the ropes is given as 2000 N.
3. Weight of the hiker: The hiker is standing on the plank, so their weight also contributes to the system.
To solve the problem, we need to use the condition that the bridge tips to an angle of 25° with respect to the horizontal. This condition tells us that the net torque acting on the system is zero. Torque is given by the formula τ = r * F * sin(θ), where r is the distance from the pivot point to the force, F is the magnitude of the force, and θ is the angle between the force and the lever arm.
In this scenario, the pivot point is the tree branch, and the forces producing torque are the weight of the plank and the weight of the hiker acting at their respective centers of mass.
Now, let's proceed with solving the problem step by step:
Step 1: Calculate the torque due to the weight of the plank.
Since the weight of the plank is acting vertically downward, it does not produce any torque about the tree branch. Therefore, the torque due to the weight of the plank is zero.
τ_plank = 0
Step 2: Calculate the torque due to the weight of the hiker.
The weight of the hiker is acting vertically downward and is located at the center of mass of the hiker (assumed to be at the midpoint of the plank). The distance from the pivot point (tree branch) to the center of mass of the hiker is L/2.
τ_hiker = r_hiker * F_hiker * sin(θ_hiker)
= (L/2) * m_hiker * g * sin(90° - θ_hiker)
= (L/2) * m_hiker * g * cos(θ_hiker)
Step 3: Calculate the maximum torque the ropes can withstand.
The maximum torque the ropes can withstand is reached when the bridge is about to tip over. At this point, the force exerted by each rope is pulling directly towards the pivot point.
τ_max = (Tension in each rope) * (Length of each rope) * sin(90°)
= (2000 N) * (2h) * 1
Step 4: Equate the torques to find the mass of the hiker.
Since the bridge is in equilibrium, the torques due to the hiker's weight and the maximum torque the ropes can withstand must balance each other.
τ_hiker = τ_max
Substituting the equations from Steps 2 and 3:
(L/2) * m_hiker * g * cos(θ_hiker) = (2000 N) * (2h)
Simplifying the equation:
(L/2) * m_hiker * g * cos(θ_hiker) = 4000 N * h
m_hiker = (4000 N * h) / (L * g * cos(θ_hiker))
Substituting the given values:
L = 2h
θ_hiker = 25°
g = 9.81 m/s² (acceleration due to gravity)
m_hiker = (4000 * 10) / (2 * 9.81 * cos(25°))
Using a calculator, we can solve for the mass of the hiker:
m_hiker ≈ 270.51 kg
Therefore, the mass of the hiker is approximately 270.51 kg.