Q59P - University Physics with Modern Physics

Question

Two workers pull horizontally on a heavy box, but one pulls twice as hard as the other. The larger pull is directed at $21.0°$ west of north, and the resultant of these two pulls is $460.0 \mathrm{N}$ directly northward. Use vector components to find the magnitude of each of these pulls and the direction of the smaller pull.

Step-by-Step Solution

Answer

The direction of smaller pull is $44.21°$ south of west, the magnitude of the larger pull is $786.41 \mathrm{N}$ and the magnitude of the smaller field is $393.21 \mathrm{N}$.

Step 1: Identification of given data

The given data is listed below,

• The angle at which a larger pull is directed is, $\theta =21°$ west to north
• The resultant pull directed northward is, $N_P=460 \mathrm{N}$.

Step 2: Concept/Significance of the vector components

The scalar coefficients are termed components when a vector is uniquely represented as a linear combination of special vectors derived from a basis.

Step 3: Determination of the magnitude of each of these pulls and the direction of the smaller pull

The resultant of two component of force is given by,

$\overrightarrow{A}+\overrightarrow{B}=\overrightarrow{C}$

Here, ${\overrightarrow{A}}_x$} is the vector component of force A, $\overrightarrow{B}$is the vector component of B and$\overrightarrow{C}$ is the resultant of two vector components.

The magnitude of the smaller pull and its direction for two unknown quantities is given by the two equations given below,

$$\begin{gathered} {\overrightarrow{A}}_x+{\overrightarrow{B}}_x={\overrightarrow{C}}_x \\ {\overrightarrow{A}}_y+{\overrightarrow{B}}_y={\overrightarrow{C}}_y \end{gathered}$$

Here, ${\overrightarrow{A}}_x$ is the horizontal vector force A, ${\overrightarrow{B}}_x$ is the horizontal vector of force B and ${\overrightarrow{C}}_x$ is the resultant of the two components of force A and B. and

The value of $B_x$ is width="110" style="max-width: none; vertical-align: -9px;" $B_x=-B\sin 21°$and $B_y$ is given by $B_y=B\sin 21°$

Let smaller pull be $\overrightarrow{A}$ and larger will be $\overrightarrow{B}$. Hence,

$B=2 A$

Substitute value in resultant formula given by,

$$\begin{gathered} A\cos \theta \hat{i}+\left(-B\sin 21°\right)\hat{i}=0 \\ 2A\sin 21°=A\cos \theta \\ \theta =44.21° \end{gathered}$$

So, the value of$\theta$ is $44.21°$.

The equation for y-direction gives values as,

$$\begin{gathered} 2A\cos 21°+A\sin \theta =460 \mathrm{N} \\ A\sin 44.21°+2A\cos 21°=460 \mathrm{N} \\ 2.5645A=460 \mathrm{N} \end{gathered}$$

So, the value of A is given by,

$A=179.37 \mathrm{N}$

$\overrightarrow{A}$must have an eastward component to cancel the westward component of$\overrightarrow{B}$ These are concluded from the figure given below,