A vector diagram shows two vectors, one of 50N pointing upwards and another vector with magnitude R at 120 degrees to the horizontal axis, with a horizontal vector of 12N. Method 1: Using cosine rule.

Physics
A vector diagram shows two vectors, one of 50N pointing upwards and another vector with magnitude R at 120 degrees to the horizontal axis, with a horizontal vector of 12N. Method 1: Using cosine rule.

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Answer

619N6\sqrt{19} N

Method 1: Using the cosine rule

Step 1: State the formula for the magnitude of the resultant force FF.

F=F12+F22+2F1F2cos⁡θF = \sqrt{F_1^2 + F_2^2 + 2 F_1 F_2 \cos \theta}

Step 2: Identify the values: F1=30 NF_1 = 30~\mathrm{N}, F2=12 NF_2 = 12~\mathrm{N}, θ=120∘\theta = 120^\circ.

Step 3: Compute cos⁡120∘\cos 120^\circ.

cos⁡120∘=−12\cos 120^\circ = -\frac{1}{2}

Step 4: Substitute into the formula.

F2=(30)2+(12)2+2(30)(12)(−12)F^2 = (30)^2 + (12)^2 + 2(30)(12)\left(-\frac{1}{2}\right)

Step 5: Calculate each term.

F2=900+144−360F^2 = 900 + 144 - 360

Step 6: Simplify.

F2=1044−360=684F^2 = 1044 - 360 = 684 F=684 NF = \sqrt{684}~N

Step 7: Simplify the square root.

684=36×19684 = 36 \times 19 F=36×19=619 NF = \sqrt{36 \times 19} = 6\sqrt{19}~N

Method 2: Using resolution of components (assuming 30 N30~\mathrm{N} vertical)

Step 1: Vertical component FyF_y.

Fy=30+12cos⁡120∘=30+12(−12)=30−6=24 NF_y = 30 + 12 \cos 120^\circ = 30 + 12\left(-\frac{1}{2}\right) = 30 - 6 = 24~N

Step 2: Horizontal component FxF_x.

Fx=12sin⁡120∘=12×32=63 NF_x = 12 \sin 120^\circ = 12 \times \frac{\sqrt{3}}{2} = 6\sqrt{3}~N

Step 3: Magnitude of resultant.

F=Fy2+Fx2=(24)2+(63)2F = \sqrt{F_y^2 + F_x^2} = \sqrt{(24)^2 + (6\sqrt{3})^2}

Step 4: Substitute and simplify.

F=576+36×3=576+108=684=619 NF = \sqrt{576 + 36 \times 3} = \sqrt{576 + 108} = \sqrt{684} = 6\sqrt{19}~N

6\sqrt{19 N}

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