Mechanics 1 Help: Solve Questions on Bead Equilibrium

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SUMMARY

The discussion focuses on solving Mechanics 1 (M1) level questions regarding a bead in equilibrium on a string. The tension in the string was initially calculated incorrectly as 10 N, but the correct tension is 7.5 N, determined by using the cosine of angle x, where tan x = 3/4. For the weight of the bead, participants confirmed that vertical force resolution is necessary, emphasizing that the sum of forces must equal zero for equilibrium.

PREREQUISITES
  • Understanding of Mechanics 1 concepts, specifically equilibrium and force resolution.
  • Familiarity with trigonometric functions, particularly sine and cosine.
  • Ability to analyze free body diagrams in physics.
  • Knowledge of tension in strings and its components in equilibrium scenarios.
NEXT STEPS
  • Study the concept of equilibrium in mechanics, focusing on static equilibrium conditions.
  • Learn how to resolve forces in two dimensions, particularly in inclined scenarios.
  • Practice drawing and analyzing free body diagrams for various mechanical systems.
  • Explore the application of trigonometric identities in solving physics problems involving angles and forces.
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Students studying Mechanics 1, physics educators, and anyone seeking to understand the principles of equilibrium and force resolution in mechanical systems.

CathyLou
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[SOLVED] Mechanics 1 help

Hi.

Could someone please help me with the following M1 level questions?

A smooth bead B is threaded on a light inextensible string. The ends of the string are attached to two fixed points A and C on the same horizontal level. The bead is held in equilibrium by a horizontal force of magnitude 6 N acting parallel to AC. The bead B is vertically below C and angle BAC = x. tan x = 3/4.

(a) Find the tension in the string.


I drew a diagram and wrote down:

Resolving horizontally:

6 - Tsinx = 0

sin x = 3/5 so T = 10 N.

However, the answer should be 7.5 N. This means that I should be using cos x but I do not understand why this is the case.

(b) Find the weight of the bead.

Would I need to resolve vertically here? If yes, could someone please help me to form the resulting equation.

I would really appreciate any help as I am really stuck at the moment.

Thank you.

Cathy
 
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For part (a), I'm assuming that you've either drawn or are given a diagram. If your diagram is set up correctly, the tension should be acting along the hypotenuse (BA) of a right angled triangle from B to A and the angle x should be the angle between the hypotenuse and the side AC. Now, you need to find the horizontal component of the tension, and this component acts parallel to the line CA which is adjacent to the angle x.

Does that make sense?
 
Hootenanny said:
For part (a), I'm assuming that you've either drawn or are given a diagram. If your diagram is set up correctly, the tension should be acting along the hypotenuse (BA) of a right angled triangle from B to A and the angle x should be the angle between the hypotenuse and the side AC. Now, you need to find the horizontal component of the tension, and this component acts parallel to the line CA which is adjacent to the angle x.

Does that make sense?

Hi.

Thank you for your help! I understand how to do part (a) now.

Could you please help with (b) also?

Cathy
 
CathyLou said:
Hi.

Thank you for your help! I understand how to do part (a) now.

Could you please help with (b) also?

Cathy
For (b), yes you have to resolve the forces vertically and you know that since the bead is in equilibrium the sum of these forces must be zero. So, what forces are acting on the bead? (Be careful with the tension)
 
Hootenanny said:
For (b), yes you have to resolve the forces vertically and you know that since the bead is in equilibrium the sum of these forces must be zero. So, what forces are acting on the bead? (Be careful with the tension)

I have worked out the correct answer now. Thank you so much for your help!

Cathy
 
CathyLou said:
I have worked out the correct answer now. Thank you so much for your help!

Cathy
A pleasure :smile:
 

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