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Math Help - Working out resultant??

  1. #1
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    Working out resultant??

    How do I work out the resultant of the following?? please see attachments hope someone can help. I know how to do them (kind of) but gone blank
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  2. #2
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    Quote Originally Posted by juangambino View Post
    How do I work out the resultant of the following?? please see attachments hope someone can help. I know how to do them (kind of) but gone blank
    For the first one, you want to put everything in x and y forces, so:

    x = 100cos(30)-25cos(40)
    y = 25sin(40)-100sin(30)

    Can you solve for x and y? To figure out the new angle, use tangent:

    tan(\frac{y}{x})=\theta
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  3. #3
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    thank you

    thank you. I already worked them out in x and y then I didnt subtract them lol must have been asleep this morning. Thank you soo much
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  4. #4
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    Quote Originally Posted by juangambino View Post
    How do I work out the resultant of the following?? please see attachments hope someone can help. I know how to do them (kind of) but gone blank
    Hello,

    to #1:

    1. I've added a sketch

    2. use cosine rule: The resulting force divides the parallelogram of forces into 2 congruent triangles. 2 legs of the triangle are the known forces and the included angle is \theta = 10^\circ. Therefore the resulting force is:

    r = \sqrt{10^2+2.5^2-2 \cdot 10 \cdot 2.5 \cdot \cos(10^\circ) }\approx 7.55 \ kN

    3. to calculate the angle between the force of 10 kN and the resulting force F_r use cosine rule again:

    \cos(\theta)=\frac{c^2-a^2-b^2}{-2 \cdot a \cdot b}.......Plug in all values you know:

    \cos(\theta)=\frac{2.5^2-10^2-7.55^2}{-2 \cdot 10 \cdot 7.55} \approx 0.998361~\implies~\theta\approx 3.281^\circ

    In my drawing I've measured the angle between the x-axis and the resulting force.
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