- #1

MrMoose

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## Homework Statement

Figure below is an overhead view of a 12kg particle that is to be pulled by three ropes. One force (F1, with magnitude 50N) is indicated. Orient the other two forces F2 and F3 so that the magnitude of the resulting acceleration of the particle is least, and find that magnitude if

(a) F2 = 30N, F3 = 20N

(b) F2 = 30N, F3 = 10N

(c) F2 = F3 = 30N.

Figure: 0----->F1

## Homework Equations

Ftx = F1 + F2x + F3x = ma

a = (F1 + F2x + F3x) / m

F2x = F2 * cos(θ)

F3x = F3 * cos(θ)

## The Attempt at a Solution

Initially, I said to myself, "well the least the acceleration can be is zero" so I assumed a = 0, Ftx = 0 and it follows that F1 = -F2x - F3x. Therefore:

F1 = -F2 * cos(θ) - F3 * cos (θ)

Solving for θ;

θ = Arcos[F1 / (-F2x - F3x)]

This gives you the angle of F2 and F3 and you've already made the assumption that a = 0

This worked out for parts (a) and (c) of the problem, but it wasn't until I dug in to part (b) that I saw a few flaws in my logic:

1) The problem doesn't say that acceleration is zero, it says to find the angle for the forces so that the acceleration is

**least**. The only problem is, how do I set up my formula so that mathematically, the difference between the sum (F2x + F3x) and F1 is as small as possible?

2) My final formula for θ assumes that both F2 and F3 have the same direction (in other words, θ2 = θ3), which may not be true. But if I introduce θ2 and θ3 in to my equations, they become very difficult to solve.

Thanks in advance, MrMoose