Normal reactions at the bases of two light supports

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Homework Help Overview

The discussion revolves around the calculation of normal reactions at the bases of two light supports in a static equilibrium scenario. The problem involves analyzing forces and torques acting on the system, particularly focusing on the relationship between the normal forces at the supports.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the application of equilibrium equations for forces and torques, with one participant presenting a detailed calculation of normal forces. Others suggest alternative methods for finding reaction forces and question the physical implications of the results, particularly regarding the relationship between the lever arms and the forces.

Discussion Status

The discussion is active, with participants exploring different approaches to the problem. Some have offered insights into the physical reasoning behind the calculated forces, while others have raised questions about the assumptions made in the problem setup, such as the nature of the supports and the conditions of equilibrium.

Contextual Notes

There are concerns regarding the assumptions of the problem, such as the presence of a true pin and the frictionless nature of the ground, which could affect the validity of the equilibrium analysis.

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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