Math  /  Trigonometry

Question5. Forces of 15 N and 11 N act a point at 125125^{\circ} to each other. Find the magnitude of the resultant.

Studdy Solution

STEP 1

1. Two forces are acting at a point.
2. The forces have magnitudes of 15N 15 \, \text{N} and 11N 11 \, \text{N} .
3. The angle between the forces is 125 125^\circ .

STEP 2

1. Recall the formula for the magnitude of the resultant of two forces.
2. Substitute the given values into the formula.
3. Calculate the magnitude of the resultant force.

STEP 3

Recall the formula for the magnitude of the resultant of two forces using the law of cosines:
R=F12+F22+2F1F2cos(θ) R = \sqrt{F_1^2 + F_2^2 + 2F_1F_2\cos(\theta)}
where R R is the resultant force, F1 F_1 and F2 F_2 are the magnitudes of the two forces, and θ \theta is the angle between them.

STEP 4

Substitute the given values into the formula:
R=152+112+2×15×11×cos(125) R = \sqrt{15^2 + 11^2 + 2 \times 15 \times 11 \times \cos(125^\circ)}

STEP 5

Calculate the magnitude of the resultant force:
First, calculate the cosine of 125 125^\circ :
cos(125)=cos(180125)=cos(55) \cos(125^\circ) = -\cos(180^\circ - 125^\circ) = -\cos(55^\circ)
Using a calculator, find cos(55)0.5736 \cos(55^\circ) \approx 0.5736 .
Thus, cos(125)0.5736 \cos(125^\circ) \approx -0.5736 .
Now, substitute back into the equation:
R=152+112+2×15×11×(0.5736) R = \sqrt{15^2 + 11^2 + 2 \times 15 \times 11 \times (-0.5736)}
R=225+121189.072 R = \sqrt{225 + 121 - 189.072}
R=156.928 R = \sqrt{156.928}
R12.52N R \approx 12.52 \, \text{N}
The magnitude of the resultant force is:
12.52N \boxed{12.52 \, \text{N}}

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