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2 33. Determine the magnitude of the resultant force and its 400 N 30P 45 800 N

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Answer #1

A more simple diagram for the given problem can be drawn as

In this diagram i have designated two force vectors as A (having magnitude 800 N)and B (having magnitude 400 N).

Now, from the figure, we can write

  overrightarrow{A} = 800 , cos , 45^{circ} , hat{i} + 800 , sin , 45^{circ} , (-hat{j})

or,  overrightarrow{A} = rac{800}{sqrt{2}} , hat{i} - rac{800}{sqrt{2}} , hat{j}

or,  A (565.685) (565.685)

And overrightarrow{B} = 400 , cos , 30^{circ} , hat{i} + 400 , sin , 30^{circ} , hat{j}

or, overrightarrow{B} = 400 imes rac{sqrt{3}}{2} ,, hat{i} + 400 , imes rac{1}{2} , , hat{j}

or. overrightarrow{B} = 346.41 ,, hat{i} + 200 , , hat{j}

So, the resultant force (F) will be

overrightarrow{F} = overrightarrow{A} + overrightarrow{B}

or, overrightarrow{F} = left ( (565.685) , hat{i} - (565.685) , hat{j} ight ) + left ( 346.41 ,, hat{i} + 200 , , hat{j} ight )

or, overrightarrow{F} = 912.09 ,, hat{i} - 365.685 , , hat{j}

Now the magnitude of resultant force (F) can be calculated as

F= left | overrightarrow{F} ight | = sqrt{(912.09)^{2} + (-365.685)^{2}} = sqrt{965633.687} = 982.67 , , N

And the direction (heta) of resultant force can be calculated as

heta = tan ^{-1} left ( rac{-365.685}{912.09} ight ) = tan ^{-1} (-0.40093) = -21.84^{circ} approx -22^{circ}

Hence, the resultant force is making an angle of nearly 220 with positive x - axis in clockwise direction.

For any doubt please comment and please give an up vote. Thank you.

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