- #1
Touran Khan
- 16
- 2
Homework Statement
Homework Equations
F = ma
The Attempt at a Solution
My value for P does not make any sense. Why is this so?
Finding the magnitude of force P
- Thread starter Touran Khan
- Start date
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- Tags
- Force Forces Magnitude Second law
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SUMMARY
The discussion focuses on calculating the magnitude of force P in a physics problem involving vertical and horizontal forces. The key equations used are F = ma, with the participant identifying the normal force (Fn) as a critical component in their calculations. The final values determined are P = 57.4 and Fn = 39.54, which satisfy the equilibrium equations for both vertical and horizontal forces. The term "reaction force" is confirmed to refer to the normal force in this context.
PREREQUISITES- Understanding of Newton's laws of motion
- Familiarity with force equilibrium equations
- Knowledge of trigonometric functions in physics
- Ability to manipulate algebraic equations
- Study the application of Newton's second law in two dimensions
- Learn about normal force calculations in various scenarios
- Explore trigonometric functions and their role in force resolution
- Investigate common mistakes in force equilibrium problems
Students studying physics, particularly those tackling mechanics and force equilibrium problems, as well as educators seeking to clarify concepts related to normal forces and vector resolution.
F = ma
My value for P does not make any sense. Why is this so?
- #2
TomHart
- 684
- 172
I think you may have forgotten one of the vertical forces.
- #3
Touran Khan
- 16
- 2
TomHart said:
Yeah, I'm almost certain that's the issue. I think it's the normal force, but what is the value of that force in this case?
- #4
TomHart
- 684
- 172
Yes, you forgot the normal force when you summed the vertical forces. But I don't know what the value of that force is. You will just have to use a variable, FN, until you solve the equations and find its value.
- #5
Touran Khan
- 16
- 2
TomHart said:
Oh right, my bad. So for vertical forces my equation ends up being Psin10 - 70cos45 + Fn = 0. From the horizontal forces equation I calculated P to be 57.369...
Then, Fn = 39.535...
To sum it up, P = 57.4 and Fn = 39.54. These values satisfy both equations, though do you have any idea what reaction force they are referring to?
- #6
TomHart
- 684
- 172
That looks right. I'm not sure but "reaction force", probably means the normal force.
- #7
Touran Khan
- 16
- 2
TomHart said:
Alright that's what I was thinking as well, no idea what else it could be. Thanks for the help.
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