Question #aecf3

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Question #aecf3

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Answer 1 · 2018-01-27T15:54:26

$20 N$ $20N$ to $45^{\circ}$ $45^@$ with the x-axis

Explanation:

Well, they are in opposite directions, and the $45^{\circ}$ $45^@$ one has more force, so it wins.

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

Since the $45^{\circ}$ $45^@$ is stronger, the resultant force with be $20 N$ $20N$ with a direction of $45^{\circ}$ $45^@$ with the x-axis.

Answer 2 · 2018-01-27T22:37:33

$39.74 N at {41.3}^{\circ}$ $39.74 N "at "41.3^@$ to the x axis.

Explanation:

I will assume that the $30^{\circ} and 45^{\circ}$ $30^@ "and 45^@$ 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
$F_{1x} = 10 N \cdot {cos 30}^{\circ} = 8.66 N$ $F_"1x" = 10 N*cos30^@ = 8.66 N$
The y axis component of the 10 N force would be
$F_{1y} = 10 N \cdot {sin 30}^{\circ} = 5.0 N$ $F_"1y" = 10 N*sin30^@ = 5.0 N$

The x axis component of the 30 N force would be
$F_{2x} = 30 N \cdot {cos 45}^{\circ} = 21.21 N$ $F_"2x" = 30 N*cos45^@ = 21.21 N$
The y axis component of the 30 N force would be
$F_{2y} = 30 N \cdot {sin 45}^{\circ} = 21.21 N$ $F_"2y" = 30 N*sin45^@ = 21.21 N$

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

The magnitude of the resultant is
$\sqrt{{(29.87 N)}^{2} + {(26.21 N)}^{2}} = 39.74 N$ $sqrt ((29.87 N)^2 + (26.21 N)^2) = 39.74 N$

The direction of the resultant is ${tan}^{- 1} (\frac{26.21}{29.87}) = {41.3}^{\circ}$ $"tan"^-1(26.21/29.87) = 41.3^@$

I hope this helps,
Steve

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Question #aecf3

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