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What is the resultant force and what is its consequent acceleration?

  • Thread starter looi76
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[SOLVED] What is the resultant force and what is its consequent acceleration?

Homework Statement


http://img48.imageshack.us/img48/4810/assignment503ij5.png [Broken]
What is the resultant force acting on the object in the figure, and what is its consequent acceleration?

Homework Equations


[tex]F = m.a[/tex]
Vertical component [tex]= F\sin\theta[/tex]
Horizontal component [tex]= F\cos\theta[/tex]

The Attempt at a Solution



http://img153.imageshack.us/img153/4963/assignment50301fq6.png [Broken]

Horizontal Component [tex]= \cos{45} \times 60 = 42.4N[/tex]
Vertical Component [tex]= \sin{45} \times 60 = 42.4N[/tex]

http://img81.imageshack.us/img81/1580/assignment50302ns9.png [Broken]

Horizontal Component [tex]= \cos{30} \times 60 = 52.0N[/tex]
Vertical Component [tex]= \sin{30} \times 60 = 30N[/tex]

Total Horizontal Component [tex]= 52 - 42 = 10N[/tex]
Total Vertical Component [tex]= 42 - (30 + 5) = 7N[/tex]

[tex]c^2 = a^2 + b^2[/tex]
[tex]x^2 = 7^2 + 10^2[/tex]
[tex]x = \sqrt{7^2 + 10^2}[/tex]
[tex]x = 12.2N[/tex]

[tex]F = m.a[/tex]

[tex]a = \frac{F}{m} = \frac{12.2}{5.0} = 2.4ms^{-2}[/tex]

Is my answer correct?
 
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Answers and Replies

  • #2
Hootenanny
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Vertical component [tex]= F\sin\theta[/tex]
Horizontal component [tex]= F\cos\theta[/tex]
You should be careful when applying these formulae, they are not universally valid. To use these formulae the angle must be measured from the horizontal. For example,
http://img153.imageshack.us/img153/4963/assignment50301fq6.png [Broken]

Horizontal Component [tex]= \cos{45} \times 60 = 42.4N[/tex]
Vertical Component [tex]= \sin{45} \times 60 = 42.4N[/tex]
This is not correct, since the angle is measured from the vertical.
http://img81.imageshack.us/img81/1580/assignment50302ns9.png [Broken]

Horizontal Component [tex]= \cos{30} \times 60 = 52.0N[/tex]
Vertical Component [tex]= \sin{30} \times 60 = 30N[/tex]
However, this solution is correct.

In general, it is better to understand how to resolve vectors rather than remembering specific formulae.
 
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  • #3
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Thnx Hootenanny, but if I measure it from the horizontal its the same because the angle is [tex]45^o[/tex] and is the final answer wrong?
 
  • #4
Hootenanny
Staff Emeritus
Science Advisor
Gold Member
9,622
6
Thnx Hootenanny, but if I measure it from the horizontal its the same because the angle is [tex]45^o[/tex] and is the final answer wrong?
Your final answers look okay to me (but I haven't checked your arithmetic), just be careful of rounding errors since you round off early in your calculations. I just wanted to emphasise that it is better to know how to result vectors rather than remembering formulae.

Secondly, remember than force and acceleration are vectors. Therefore, you need to quote the direction is which they act.
 

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