Question

Help pls i need this right now

Answer 1

The x-component of
`F_(3)` is 56.148 newtons.

Step-by-step explanation:

From 1st and 2nd Newton's Law we know that a system is at rest when net acceleration is zero. Then, the vectorial sum of the three forces must be equal to zero. That is:

`\vec F_(1) + \vec F_(2) + \vec F_(3) = \vec O` (1)

Where:

`\vec F_(1)`,
`\vec F_(2)`,
`\vec F_(3)` - External forces exerted on the ring, measured in newtons.

`\vec O` - Vector zero, measured in newtons.

If we know that
`\vec F_(1) = (70.711,70.711)\,[N]`,
`\vec F_(2) = (-126.859, 46.173)\,[N]`,
`F_(3) = (F_(3,x),F_(3,y))` and
`\vec O = (0,0)\,[N]`, then we construct the following system of linear equations:

`\Sigma F_(x) = 70.711\,N - 126.859\,N +F_(3,x) = 0\,N` (2)

`\Sigma F_(y) = 70.711\,N + 46.173\,N+F_(3,y) = 0\,N` (3)

The solution of this system is:

`F_(3,x) = 56.148\,N`,
`F_(3,y) = -116.884\,N`

The x-component of
`F_(3)` is 56.148 newtons.