Question

Question #aecf3

Answer 1

`20N` to `45^@` with the x-axis

Explanation:

Well, they are in opposite directions, and the `45^@` one has more force, so it wins.

Thus, the resultant force is `30N-10N=20N`.

Since the `45^@` is stronger, the resultant force with be `20N` with a direction of `45^@` with the x-axis.

Answer 2

`39.74 N "at "41.3^@` to the x axis.

Explanation:

I will assume that the `30^@ "and 45^@` angles are both counterclockwise from the +x axis. (Or both could be clockwise -- my result would not be different if so.)

The x axis component of the 10 N force would be
`F_"1x" = 10 N*cos30^@ = 8.66 N`
The y axis component of the 10 N force would be
`F_"1y" = 10 N*sin30^@ = 5.0 N`

The x axis component of the 30 N force would be
`F_"2x" = 30 N*cos45^@ = 21.21 N`
The y axis component of the 30 N force would be
`F_"2y" = 30 N*sin45^@ = 21.21 N`

The x component of the resultant is
`8.66 N +21.21 N = 29.87 N`
The y component of the resultant is
`5.0 N +21.21 N = 26.21 N`

The magnitude of the resultant is
`sqrt ((29.87 N)^2 + (26.21 N)^2) = 39.74 N`

The direction of the resultant is `"tan"^-1(26.21/29.87) = 41.3^@`

I hope this helps,
Steve