SOLUTION: okay Im trying to help my daughter we got some answers but want to make sure. Bob excerts 30.0-n force to the left on a box (m=100.0kg) Carol excerts 2

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Question 517483: okay Im trying to help my daughter we got some answers but want to make sure.
Bob excerts 30.0-n force to the left on a box (m=100.0kg)
Carol excerts 20.0-n force on the same box perpendicular to Bobs force whats the net force on the box and determine the acceration of the box and at what rate would the box accerate if both forces where to the left instead of perpendicular to each other? thank you.

Answer by bucky(2189) (Show Source):
You can put this solution on YOUR website!
Thank you for helping your daughter to better understand the world she lives in.
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"Fysics is fun" ... or is it "Physics is Phun"?
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Anyhow, in the problem you can ignore friction because it isn't a part of the given information. The way I would start this is to draw a horizontal base line and sketch a rectangle representing the box and using the extended base line as the bottom side of the box. Then put a dot in the center of the box (where the diagonals of the box cross).
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Label the box 100 kg. Starting at the center dot draw a horizontal vector line to the left and label it 30 newtons. Then return the dot and starting there draw a vertical vector and label it 20 newtons. (Just to keep things sort of to scale, make the length of this vector about two-thirds the length of the first vector of 30 newtons.)
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Start at the arrowhead point at the end of the 20 newton vector and extend a horizontal line to the left (parallel to the 30 newton vector that you first drew.) Now go to the arrowhead point on the original 30 newton vector and starting there draw a vertical line upward and parallel to the original 20 newton vector. Extend this vertical line until it intersects the horizontal line coming from the arrowhead. You have now formed what is called the force parallelogram, in this case a rectangle having two sides formed by the original 30 newton and 20 newton vectors and the other two sides (meeting at an intersecting point) formed by the vertical line through the left end of the 30 newton vector and the horizontal line through the top of the 20 newton vector arrowhead.
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I hope this is understandable. Now draw a new vector that starts at the center dot of the box and goes left and upward to the intersecting point of the lines from the ends of the two original vectors. Put the arrowhead at the end of this line where it meets the intersecting point. Now we're ready to go to work. This new vector (dot to intersecting point) represents the force (both direction and magnitude) that results from the 30 newton vector and the 20 newton vectors applied to the box.
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You can now apply the Pythagorean theorem to calculate the magnitude of the force represented by this new vector. The new vector is the hypotenuse of the right triangle with legs of 20 newtons and 30 newtons. By the Pythagorean theorem the hypotenuse (Resultant Force R) is:
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and the resultant force R is the square root of 1300 or:
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newtons.
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This force is equal to the Mass (100 kg) times the Acceleration. So we can write:
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R = M*A
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36.0555 = 100 * A
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Solving for the acceleration we get:
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A = 36.0555/100
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which gives an answer of 0.360555 meters/sec^2 for the acceleration in the direction of the Resultant force vector.
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Instead of the Pythagorean theorem you could have also done this by trigonometry. It's time to use a little trig to find one of the angles that helps to identify the direction of the vector. (I hope your daughter has had some trig so this isn't overwhelming.) You can get the angle between the 30 newton vector and the Resultant force vector by using the arctangent function. For this function the 20 newton side is the opposite side (call it O) and the 30 newton side is the adjacent side (call it A). We write:
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You can use a calculator to find that this angle is 33.69 degrees. Again, this is the angle between the 30 newton vector and the Resultant force vector.
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If both forces act together on the box and both are to the left, the resulting combined force is 30 + 20 = 50 newtons. Again using the formula that Force equals Mass times Acceleration we get:
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Divide both sides by 100 and get:
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which reduces to the acceleration A = 0.5 meters/sec^2.
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Check my math to ensure that I didn't make a dumb mistake somewhere along the line.
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I hope these answers match yours, and that they may help you to understand the problem a little better.
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