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Chapter 11: Problem 74

A uniform, \(90.0-\mathrm{N}\) table is 3.6 \(\mathrm{m}\) long, 1.0 \(\mathrm{m}\) high, and 1.2 \(\mathrm{m}\) wide. \(\mathrm{A} 1500\) -N weight is placed 0.50 \(\mathrm{m}\) from one end of the table, a distance of 0.60 \(\mathrm{m}\) from each side of the table. Draw a free body diagram for the table and find the force that each of the four legs exerts on the floor.

Short Answer

Expert verified
Forces on legs: F1=F2=690.835 N, F3=F4=104.17 N.

Step by step solution

01

Understanding the Problem

First, we need to understand the setup. The table is supported by four legs at the corners. A 1500-N weight is placed 0.5 m from one end of the table. We must determine how the weight distribution affects the force on each leg.
02

Analyzing the Table's Weight

The total weight of the table is 90.0 N. Being uniform, the weight is evenly distributed among all four legs. Without any additional weight, each leg supports a weight of \( \frac{90.0}{4} = 22.5 \mathrm{N} \).
03

Analyzing the External Weight's Position

The 1500-N weight is positioned asymmetrically, 0.5 m from one end along the length (3.6 m) and centered across the width (1.2 m). This affects the distribution of forces among the legs.
04

Drawing the Free Body Diagram

In the free body diagram, show the four forces \( F_1, F_2, F_3, \) and \( F_4 \) exerted by each leg on the floor. The gravitational force of 90.0 N acts at the center, and the 1500-N force acts 0.5 m from the left end of the table.
05

Applying Static Equilibrium Conditions

For equilibrium, the sum of vertical forces and torques about any point should be zero. Let \( F1 \) and \( F2 \) be the forces on the left legs, and \( \sum F_y = 0 \), gives:\[ F_1 + F_2 + F_3 + F_4 = 90.0 + 1500 \]\[ F_1 + F_2 + F_3 + F_4 = 1590 \mathrm{N} \]
06

Calculating Torque Relative to Leftmost End

Find torques about the left end:Torque by 1500-N load is \( 1500 \times 0.5 \), counter-clockwise.Balance torques by right legs:\[ F_3 \times 3.6 + F_4 \times 3.6 = 1500 \times 0.5 \]\[ (F_3 + F_4) \times 3.6 = 750 \]\[ F_3 + F_4 = \frac{750}{3.6} \approx 208.33 \mathrm{N} \]
07

Solving for Forces on the Left Side

Using the equilibrium of vertical forces:\[ F_1 + F_2 + 208.33 = 1590 \F_1 + F_2 = 1590 - 208.33 \F_1 + F_2 = 1381.67 \mathrm{N} \]
08

Distributing Unevenly based on Symmetry

Assuming symmetry on both width axes and considering 0.6 m from sides, forces should be distributed equally across pairs:\( F_3 = F_4 = 104.17 \mathrm{N} \)Similarly, redistribute left side pairs:\( F_1 = F_2 = 690.835 \mathrm{N} \)
09

Verification

Check that sum \( F_1 + F_2 + F_3 + F_4 = 690.835 \times 2 + 104.17 \times 2 \approx 1590 \mathrm{N} \). This satisfies all equilibrium conditions.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Free Body Diagram
To analyze static equilibrium problems, a Free Body Diagram (FBD) is essential. It helps visualize all forces acting on an object, like the table in our problem. This diagram is a simple representation using arrows to indicate force direction and magnitude.

For our table, the diagram includes:
  • Four force vectors at each table leg: \( F_1, F_2, F_3, \) and \( F_4 \).
  • A downward force at the table鈥檚 center due to its weight (90.0 N).
  • A second downward force positioned 0.5 m from the table鈥檚 end - the 1500 N weight.
The FBD serves as a visual tool, simplifying the use of equations to solve for unknown forces. Drawing a clear, accurate diagram is crucial before proceeding with calculations.
Torque Calculation
Torque, or the rotational force, plays a vital role when analyzing equilibrium. It's calculated using the formula \( \text{Torque} = \text{Force} \times \text{Distance} \). In our scenario, torque impacts how the additional 1500 N weight affects the table legs.
  • The torque caused by the weight is calculated as \( 1500 \times 0.5 = 750 \) Nm, acting counter-clockwise around the leftmost end of the table.
Balancing torque involves ensuring torques about a pivot equal zero. Thus, the opposing torques created by the forces at the right legs \( F_3 \) and \( F_4 \) must counteract this. Mathematically:
  • \( (F_3 + F_4) \times 3.6 = 750 \) leading to \( F_3 + F_4 \approx 208.33 \) N.
This balance ensures the table doesn鈥檛 rotate, maintaining static equilibrium.
Weight Distribution
Understanding weight distribution is crucial in determining the forces that table legs must handle. In static equilibrium, both the table's weight and the additional 1500 N weight need consideration.
  • The table's own weight (90.0 N) is uniformly distributed across its four legs, with each leg initially supporting 22.5 N.
  • The 1500 N weight, however, is not centered, causing uneven force distribution.
  • Its position affects the left and right side legs differently, necessitating adjustments in calculating forces \( F_1, F_2, F_3, \) and \( F_4 \).
Through analysis and calculation, we found \( F_3 = F_4 = 104.17 \) N and \( F_1 = F_2 = 690.835 \) N after the account of extra load. Accurate weight distribution ensures all forces are balanced properly.
Equilibrium Conditions
In static equilibrium, an object experiences no net force or torque. For our table:
  • The sum of all vertical forces \( F_1, F_2, F_3, \) and \( F_4 \) must equal the total downward forces (90.0 N + 1500 N = 1590 N).
  • The torques about any pivot, like the table's leftmost end, must also sum to zero.
Applying these conditions, equations were set for sums of forces and torques. Solving them ensured balance:
  • The forces on the left \( F_1 + F_2 \) resulted in 1381.67 N.
  • The forces on the right \( F_3 + F_4 \) totalled 208.33 N.
  • Verification checks ensured that all forces and torques met the equilibrium constraints, with the sum equalling 1590 N overall.
Adhering to these principles is key to solving similar static equilibrium challenges correctly.

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