Chapter 9: Problem 22
(II) A 20.0-m-long uniform beam weighing 650 N rests on walls A and B, as shown in Fig. 9-62. (a) Find the maximum weight of a person who can walk to the extreme end D without tipping the beam. Find the forces that the walls A and B exert on the beam when the person is standing: (b) at D; (c) 2.0 m to the right of A.
Short Answer
Step by step solution
Understanding the Problem
Define the Coordinates and Distances
Set Up Conditions for Non-Tipping
Calculate Maximum Person's Weight
Force Calculation at Point D
Summing Torques About A When Person is at D
Force Calculation at 2m Right of A
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Torque calculation
The basic formula for torque is: \[\tau = r \times F \times \sin(\theta),\]where
- \(\tau\) is the torque,
- \(r\) is the distance from the pivot point to the point where the force is applied,
- \(F\) is the force applied,
- \(\theta\) is the angle between the force vector and the lever arm.
Beam balance
There are a few elements at play:
- A beam with its weight acting at its center of gravity.
- A person moving along the beam adding additional weight.
- The supports at walls A and B exerting reactionary forces.
Forces and moments
To analyze our beam problem, we calculate
- All the forces acting vertically, like the weight of the beam and the person walking on it.
- The reaction forces at the points of support.
- Moments created by these forces about a pivot point (either end or within the beam).
Non-tipping condition
When a person walks on the beam:
- Their weight adds a new torque component.
- The total torque about either support point must remain zero to avoid tipping.
This balance prevents any rotation, maintaining stability and satisfying the non-tipping condition.