Normal reactions at the bases of two light supports

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SUMMARY

The discussion centers on calculating the normal reactions at the bases of two light supports under equilibrium conditions. Using the equations of static equilibrium, the normal forces were determined to be ##n_1 = 108 \; \text{N}## and ##n_2 = 192 \; \text{N}##. The relationship between the two forces was established as ##n_2 = \frac{16}{9} n_1##, indicating that ##n_2## is greater than ##n_1## due to its shorter lever arm, which compensates for the torque produced by ##n_1##. The discussion also highlights alternative methods for solving the problem, emphasizing the importance of understanding the physical implications of the results.

PREREQUISITES
  • Understanding of static equilibrium principles
  • Familiarity with normal force calculations
  • Knowledge of torque and lever arm concepts
  • Basic proficiency in trigonometry
NEXT STEPS
  • Study the principles of static equilibrium in mechanics
  • Learn about torque calculations and their applications
  • Explore different methods for solving equilibrium problems
  • Investigate the effects of friction on support systems
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brotherbobby
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Homework Statement
Shown in the figure below are two light plastic supports pinned at their top. One of the supports is 4 m long while the other is 3 m long and they are arranged to form a right angled triangle with their base 5 m. A mass of 300 N is hung from the pin at the top. If the ground is frictionless, calculate the reactions ##n_1## and ##n_2## at the bases of the two supports.
Relevant Equations
For equilibrium, ##\Sigma \vec F = 0## for the system as a whole and ##\Sigma \vec \tau = 0## for any point on the system where torque ##\vec \tau = \vec r \times \vec F##.
reactions.png


For equilibrium, using ##\Sigma \vec F = 0##, we get ##n_1 + n_2 = 300\; \text{N}##.

Taking the system as a whole and applying ##\Sigma \vec \tau = 0## about the hinge (pin) at the top from where the load is hung, we get ##n_1 \times (0.8) \times 4 = n_2 \times (0.6) \times 3##, by taking those components of the normal forces perpendicular to the supports and using trigonometry.

Hence, ##3.2 n_1 = 1.8 n_2 \Rightarrow n_2 = \frac{16}{9} n_1##.

Thus, going to the earlier equation, ##n_1 + \frac{16}{9} n_1 = 300 \Rightarrow \frac{25}{9} n_1 = 300 \Rightarrow \boxed{n_1 = 108 \; \text{N}}##.
Also, ##\boxed{n_2 = 192 \; \text{N}}##.

Is this right?

Even if it is, is there a physical explanation as to why ##n_2 > n_1##? Can we say it is because ##n_2## has a smaller lever arm than ##n_1## and therefore has to be greater in order to nullify the torque produced by ##n_1## about the hinge?
 
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Or that the mass of the "object" is horizontally closer to point 2 which therefore sees more force. There are a hundred ways to say it! Very nicely done.
 
Another approach is to take moments about each support point in turn and dispense with the vertical force balance equation. This has the slight advantage of avoiding having to solve a pair of simultaneous equations, but it is more useful when you only need to find one reaction force.
 
Where did this problem come from? If there is a true pin at the Top, and if the supports at their base sit on frictionless ground, this system will instantly collapse.
 

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