during a three-way tug-of-war competition, each of the three ropes is tied to a center knot. one man pulls his rope north with a force of 301 n. another man pulls his rope at 10.0° south of east with a force of 280 n. if the third man pulls on his rope southwest with a force of 313 n, what is the net force on the knot?

62.7 n
42.7 n
72.7 n
52.7 n

Question

during a three-way tug-of-war competition, each of the three ropes is tied to a center knot. one man pulls his rope north with a force of 301 n. another man pulls his rope at 10.0° south of east with a force of 280 n. if the third man pulls on his rope southwest with a force of 313 n, what is the net force on the knot?

62.7 n
42.7 n
72.7 n
52.7 n

Accepted Answer

{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "multiple_choice_formatter", "explain_then_answer", "tutor", "mastery_adaptive_brevity" ], "needs_drawing": true, "concepts_used": [ "Vector Resolution", "Net Force Calculation" ], "new_concepts": [], "current_concepts": [ "Vector Resolution", "Net Force Calculation" ] } ## Define the coordinate system and given forces Using the **Vector Resolution** and **Net Force Calculation** knowledge points $$ \begin{aligned} &\text{Let East be the positive } x\text{-axis and North be the positive } y\text{-axis.}\ &\vec{F}_1 = 301\text{ N North} \implies \vec{F}_1 = (0)\hat{i} + (301)\hat{j}\ &\vec{F}_2 = 280\text{ N at } 10.0^\circ\text{ South of East} \implies \vec{F}_2 = (280\cos(-10.0^\circ))\hat{i} + (280\sin(-10.0^\circ))\hat{j}\ &\vec{F}_3 = 313\text{ N Southwest } (45^\circ\text{ South of West}) \implies \vec{F}_3 = (-313\cos(45^\circ))\hat{i} + (-313\sin(45^\circ))\hat{j} \end{aligned} $$ ## Resolve forces into components Using the **Vector Resolution** knowledge point $$ \begin{aligned} &F_{1x} = 0\text{ N},\quad F_{1y} = 301\text{ N}\ &F_{2x} = 280 \times 0.9848 = 275.74\text{ N},\quad F_{2y} = 280 \times (-0.1736) = -48.62\text{ N}\ &F_{3x} = -313 \times 0.7071 = -221.33\text{ N},\quad F_{3y} = -313 \times 0.7071 = -221.33\text{ N} \end{aligned} $$ ## Calculate net force components Using the **Net Force Calculation** knowledge point $$ \begin{aligned} &F_{\text{net}, x} = F_{1x} + F_{2x} + F_{3x} = 0 + 275.74 - 221.33 = 54.41\text{ N}\ &F_{\text{net}, y} = F_{1y} + F_{2y} + F_{3y} = 301 - 48.62 - 221.33 = 31.05\text{ N} \end{aligned} $$ ## Calculate net force magnitude Using the **Net Force Calculation** knowledge point $$ \begin{aligned} &F_{\text{net}} = \sqrt{F_{\text{net}, x}^2 + F_{\text{net}, y}^2}\ &F_{\text{net}} = \sqrt{(54.41)^2 + (31.05)^2} = \sqrt{2960.45 + 964.10} = \sqrt{3924.55} \approx 62.65\text{ N} \end{aligned} $$

(A) 62.7 N

(B) 42.7 N

(C) 72.7 N

(D) 52.7 N

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QUESTION IMAGE

Question

Explanation:

Define the coordinate system and given forces

Resolve forces into components

Calculate net force components

Calculate net force magnitude

Answer: