Understanding Forces and Vectors: Solving a 2D Acceleration Problem

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Homework Help Overview

The discussion revolves around a physics problem involving two forces acting on a 3.00 kg object, specifically focusing on calculating the magnitude and direction of the resulting acceleration in a two-dimensional context. The forces include a 40.0 N force in the x direction and a 60.0 N force at a 45-degree angle, leading to questions about vector components and resultant forces.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss breaking down the forces into their x and y components and adding these components to find the resultant force. There are questions about the correctness of the calculations and the use of trigonometric functions, particularly regarding angles and the cosine rule.

Discussion Status

The discussion includes various attempts to clarify the calculations, with some participants suggesting different methods to arrive at the acceleration. There is recognition of discrepancies between personal calculations and the book's answers, prompting further exploration of the problem setup and assumptions.

Contextual Notes

Some participants express confusion over the use of trigonometric values and the interpretation of the problem, indicating a need for clarification on the setup and the forces involved. There is also mention of specific numerical values that differ from expected results, highlighting potential misunderstandings in the calculations.

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First here is the problem: Only two forces act on an object(mass 3.00kg) as in the drawing(40.0N in the x direction and 60.0N north-east and a 45 degree angle between them) Find the magnitude and direction, relative to the x-axis of the acceleration of the object.

I have tried finding the x and y components of each vector and then adding the x components and y components together to get the resultant and then using a=F/M and the tangent to determine the angle(direction) BUT I keep getting the wrong answer! Please explain!
 
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Did you get this for the components : first force = 40e_x
second force = 60(cos(45)e_x + sin(45)e_y) and cos(45) = sin (45)= 1/sqrt(2)?

e_x and e_y just denote the basis-vectors...

If you did get this, you are right...

Remeber that is this exercise you only need to add the x-components in order to find the answer. We are only looking in the x-direction here..
 
Shouldn't the answer be 27,47546896 m/s² ?
 
That is how I did the problem, but the answer in the book is a=30.9(m/s2) and the degree is 27.2 degrees above the x-axis??!
 
Ahh, got it

They are asking the resulting force when you add the two given forces together

use the cosinus-rule...
 
F² = 40² + 60² -2*40*60*cos(135) Then take the squareroot and divide by three and you will get 30,9 m/s²
 
Or with the components

(total force)² = (60/sqrt(2) + 40)² + (60/sqrt(2))² = 92,7

The angle is calculated using tan(x)=Y/X and Y component = 60/sqrt(2)
and X component = (40 + 60/sqrt(2))

regards
marlon
 
Do what marlon says or

[tex]R_{x} = 60cos(45) + 40cos(0)[/tex]

[tex]R_{y} = 60sin(45) + 40sin(0)[/tex]

[tex]\vec{F_{R}} = (82.42i + 42.42j) N[/tex]

[tex]\vec{A} = (\frac{82.42}{3}i + \frac{42.42}{3}j) m/s^2[/tex]

[tex]\vec{A} = (27.473i + \14.14j) m/s^2[/tex]

[tex]|\vec{A}| = \sqrt{27.473^2+14.14^2}[/tex]

[tex]|\vec{A}| = 30.89[/tex]

[tex]\theta_{A} = arctan(\frac{14.14}{27.473})[/tex]

[tex]\theta_{A} = 27.23^o[/tex]
 
Thanks! I understand I was using cos(45) not cos(0)!
 

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