Need major help w/ this physics problem, finding the magnitude and theta?

NetherCraft 0

At a picnic, there is a contest in which hoses are used to shoot water at a beach ball from three directions. As a result, three forces act on the ball, F1, F2, and F3 (see drawing). The magitudes of F1 and F2 are F1 = 50.0 newtons and F2 = 90.0 newtons. Using a scale drawing and the graphical technique determine the following such that the resultant force acting on the ball is zero.

here is link to drawing http://i48.photobucket.com/albums/f226/buttrefly18…

2 Answers

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    okay so, what you do is, u see that F2 is right on the x-axis, so you can just leave it as is. but you see how F1 is a force at some angle? well, you have to break that F2, which is the “resultant force” back down into its’ x and y components. you do that with ur simple trig functions.

    F1=50N

    lets find the x component first.

    cos(θ)=Adj/Hyp

    basically you just plug stuff in and solve for the adj, you know hyp you know θ, so just plug n solve.

    cos(60)50=Adj=25

    btw, when ur doing this in ur calculator, remember that you must do close parenthesis after the angle 60, this is because “sin, cos, tan” are trig FUNCTIONS, so even though it looks like ur multiplying 60×50, ur not, ur actually multiplying 50 x {cos(60)}, the “cosine OF 60”.

    so….

    Adj=25

    now lets find y

    sin(60)50=Op=43.30127

    and how much force remaining in the x and y components?

    ΣF(x)=90-25=65

    so this means that the x component of F3 must have a magnitude of 65N.

    ΣF(y)=43.30127, because this is the only y acting force, all of F2 is in the x direction

    this means that the y component of F3 must have a magnitude of 43.30127N.

    if you just use ur pythagorian therom, and solve for the resultant, or.. hypotenuse, you get about 78.1N, and what is this angle? this angle is the arc tangent of 43.30127019 / 65.

    the picture shows this angle being in respect to the x-axis.

    and if you don’t know what arc tangent is,

    arc tangent = “{tan^(-1)}(θ)”, basically tangent of theta, “tanθ” except, the “tangent” is to the negative one. we call this arc tangent. remember we were solving for force components of individual legs of the triangle in F1? well, you use arc tangent when you’re trying to solve for the angle. then you do arc tangent. basically, u noe how “^-1” means ur inverting something? which means ur dividing by one, so ur basically dividing both sides by “tangent”, and ur solving for theta.

    tangent^-1(θ)=(43.30127019 / 65)= (about) 33.67*

    =)

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