Forces F(40N, 045°) and Q(PN, 135°) act on a body initially at rest. If the magnitude of their resultant is 50N, find the value of P
Asked
Active
Viewed 26 times
-3
-
It's a degree sign – Saeed Mar 21 '17 at 00:21
-
I mean, is it $45^{\circ}$? Its not clear because $045^{\circ}$ is not significant. – Juniven Acapulco Mar 21 '17 at 00:25
-
Oh yeah it is 45° – Saeed Mar 21 '17 at 00:28
1 Answers
0
Break up your two forces into their $x$ and $y$ components. To get the $x$ component, multiply the magnitude of your force by the cosine of your angle, and to get the $y$ component, multiply the magnitude of your force by the sine of your angle.
$F=(F_x)\hat{x}+(F_y)\hat{y}=(40\cos(45^\circ))\hat{x}+(40\sin(45^\circ))\hat{y}$
$Q=(Q_x)\hat{x}+(Q_y)\hat{y}=(P\cos(135^\circ))\hat{x}+(P\sin(135^\circ))\hat{y}$
You know the magnitude of their resultant vector is $50$N, so that means $$50=\sqrt{(F_{x}+Q_{x})^2+(F_{y}+Q_{y})^2}$$
Can you plug in the corresponding values and solve for P?
Jonathan Barkey
- 111
- 10
-
-
@Saeed, no problem! If you found that this answer was satisfactory to your question, don't forget you can mark this as the accepted answer by clicking the check mark. – Jonathan Barkey Mar 21 '17 at 00:35