At a picnic, there is a contest in which hoses are used to shoot water at a beac

Question

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 . The magnitudes of F1 and F2 are F1 ?? 50.0 newtons and F2 ?? 90.0 newtons. determine (a) the magnitude of f3 (b) the angle q such that the resultant force acting on the ball is zero question-in this problem you know to subtract 25 from 90 in order to get the x value of the resultant(65) because the angle between the two vectors form an angle that is greater than 90 degrees ? or if one wants to find the x value or y value of any resultant vector you simply subtract their x and y values of the separate components

Explanation / Answer

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(?)=Adj/Hyp basically you just plug stuff in and solve for the adj, you know hyp you know ?, 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 60x50, 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? SF(x)=90-25=65 so this means that the x component of F3 must have a magnitude of 65N. SF(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)}(?)", basically tangent of theta, "tan?" 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(?)=(43.30127019 / 65)= (about) 33.67*

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