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How to find Normal force?

by texasgirl03
Tags: force, normal
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texasgirl03
#1
Nov4-09, 11:14 PM
P: 3
1. The problem statement, all variables and given/known data

If a box that weighs 536N is pulled forward at a constant speed by a force of 150N at an angle of 37 degreees with the ground, what normal force does thte box exert on the supporting surface?

( I am really not sure at all where to begin) formula?
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cepheid
#2
Nov4-09, 11:35 PM
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Hi texasgirl03,

You know the magnitude of the force pulling on the box and its angle from the horizontal. So, you can resolve this force into two components, one acting vertically, and the other acting horizontally. That's the starting point.

Do you know what a free body diagram is? It would be helpful to draw one here.
texasgirl03
#3
Nov4-09, 11:44 PM
P: 3
Okay i understand so far.. but what is the formula?

cepheid
#4
Nov4-09, 11:56 PM
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How to find Normal force?

If you draw a triangle with the force at 37 degrees from the horizontal, you can use basic trigonometry to get the horizontal and vertical components of that force (the trigonometric ratios, sine and cosine, are what will be used). Draw the triangle that resolves the force vector into its components and it will be clear.
texasgirl03
#5
Nov4-09, 11:59 PM
P: 3
I do not know how to do that. thats why i am asking. I just wanted the formula.
cepheid
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Nov5-09, 12:06 AM
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Quote Quote by texasgirl03 View Post
I do not know how to do that. thats why i am asking. I just wanted the formula.
No disrespect, but drawing a right-angled triangle and applying basic trigonometry are both things that you should know how to do, and if I just give you the answer, you won't really learn much.

Draw the force vector at an angle of 37 degrees from the horizontal. Now, you can see that this vector can be represented as the sum of two other component vectors, a horizontal component, and a vertical component. Together these two components add up in the usual way for vector addition to produce the force vector. These three vectors form a right-angled triangle. The ratio of the side of the triangle that is opposite to the angle to the hypotenuse is the sine of that angle. The ratio of the adjacent side and hypotenuse is the cosine of the angle. That is what you need to know in order to calculate the horizontal and vertical components (but I am not going to tell you which one is which).
Deep_Blue
#7
Nov5-09, 01:32 AM
P: 2
Hi Texasgirl03,

Take a look a the diagram and see if it makes the problem any clearer.
Notice that there's two forces acting upwards the Normal force (N) & the y component of the 150N force (Ty), and one force pulling downwards (Fg). Find Ty and remember that according to Newton's Second Law (∑ F = m*a), the sum of all forces in the y direction will equal the object's mass * acceleration. See if you can figure it out from here.



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