- #1

- 17

- 2

- Homework Statement
- 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 degrees west of north, and the resultant of these two pulls is 460.0N directly northward. Use vector components to find the magnitude of each of these pulls and the direction of the smaller pull.

- Relevant Equations
- ##A_x+B_x = C_x##

My solution for finding the direction of the smaller pull is slightly off from the text solution. I am unsure why.

Assuming the larger vector is ##\vec B##, the smaller ##\vec A##, and the resultant ##\vec C##. My solution for the direction of this smaller pull (for the smaller pull in quadrant one) is as follows:

My solution:

##A_x+B_x = C_x##

##| \vec A |cos \phi + 2| \vec A | cos111° = 0 ##

##| \vec A |(cos \phi + 2cos111°) = 0##

From zero product property

##cos \phi + 2cos111° = 0##

## \phi = 44.21°##

Text book solution:

##2| \vec A |sin21.0° = | \vec A |sin\phi##

## \phi = 45.79°##

I do not understand why there is a difference.

Assuming the larger vector is ##\vec B##, the smaller ##\vec A##, and the resultant ##\vec C##. My solution for the direction of this smaller pull (for the smaller pull in quadrant one) is as follows:

My solution:

##A_x+B_x = C_x##

##| \vec A |cos \phi + 2| \vec A | cos111° = 0 ##

##| \vec A |(cos \phi + 2cos111°) = 0##

From zero product property

##cos \phi + 2cos111° = 0##

## \phi = 44.21°##

Text book solution:

##2| \vec A |sin21.0° = | \vec A |sin\phi##

## \phi = 45.79°##

I do not understand why there is a difference.