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[SOLVED] What is the resultant force and what is its consequent acceleration?
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?
[tex]F = m.a[/tex]
Vertical component [tex]= F\sin\theta[/tex]
Horizontal component [tex]= F\cos\theta[/tex]
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?
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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