# Two dogs pull horizontally on ropes attached to a post; the angle between the ropes is 60.0...

## Question:

Two dogs pull horizontally on ropes attached to a post; the angle between the ropes is 60.0 degrees. If dog A exerts a force of 270N and dog B exerts a force of 300N,

find the magnitude of the resultant force and the angle it makes with dog A's rope.

## Vector Addition:

This problem can be solved using the parallelogram method and this method gives the rule for vector addition of two or more vectors. It is done graphically by placing vector A and B at the same initial point then complete the parallelogram shape.

## Answer and Explanation:

Given:

- Angle between the ropes : {eq}\theta = 60^{\circ}{/eq}

- Force exerted by Dog A : {eq}F_A = 270\ N{/eq}

- Force exerted by Dog B : {eq}F_B = 300\ N{/eq}

Find:

- Magnitude and direction of the resultant force : {eq}F_R =?{/eq} and {eq}\theta = ?{/eq}

To further understand the question we need to visualize it first using the figure as shown below,

**(Part 1: Magnitude)**

We will use the parallelogram Law of vector addition to solve for the resultant force

$$\begin{align} F_R^2 &= F_A^2 + 2F_A F_B \cos \theta +F_B^2 \\[0.3cm] F_R &= \sqrt {(270\ \rm N)^2 + 2 (270\ \rm N)(300\ \rm N)(cos(60^{\circ})) + (300\ \rm N)^2) } \\[0.3cm] F_R &= \sqrt{243900 \ \rm N^2} \\[0.3cm] F_R &\approx \boxed{493.9\ \rm N} \end{align} $$

**(Part 2: Direction)**

To get the direction, we will use the formula,

$$\begin{align} \theta &= \tan^{-1}\left(\frac{\sum y}{\sum x} \right) \\[0.3cm] &= \tan^{-1} \left(\frac{F_B \sin \theta}{F_A +F_B \cos \theta} \right) \\[0.3cm] &= \tan^{-1} \left[ \frac{(300\ \rm N) \sin(60^{\circ})}{(270\ \rm N)+(300\ \rm N) \cos(60^{\circ})} \right] \\[0.3cm] &= \tan^{-1} \left(\frac{259.8}{420} \right) \\[0.3cm] &= \tan^{-1}(0.62^{\circ}) \\[0.3cm] &\approx \boxed{31.7^{\circ}} \end{align} $$

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from UExcel Physics: Study Guide & Test Prep

Chapter 2 / Lesson 2