Question

Two forces acting on a particle. Force A is 3.0 N directed 

north and the other force B is 4.0 N 300 N of E. Find the 

resultant force using Tail-to-Tip, parallelogram, and 

component method.

Answer 1

Explanations & Calculations

• Since the exact direction of the second force is not given, consider the figure attached which describes the 2 possible situations.

LET P= 3N , Q = 4N

• When the parallelogram theorem to be used, you can either draw the force arrangement to some scale & measure the length of the line segment for the resultant force.
• Or you can employ the equation meant for the parallelogram theorem

$\qquad\qquad \begin{aligned} \small R&=\small \sqrt{P^2+Q^2+2pq\cos\theta} \end{aligned}$

• By that, for both the situations

$\qquad\qquad \begin{aligned} \small R &=\small \sqrt{3^2+4^2+2.(3).(4)\cos60}\\ &=\small \sqrt{37}\approx6.0828\,N\\\\ or\\\\ \small R_1&=\small \sqrt{3^2+4^2+2(3)(4)\cos(90+30)}\\ &=\small \sqrt{13}\approx3.6056\,N \end{aligned}$

• To evaluate the resultant by component method, consider the total X and Y components as shown.
• Since those 2 components (resolutions) are orthogonal, the resultant is given by Pythagoras theorem,

$\qquad\qquad \begin{aligned} \small R&=\small \sqrt{X^2+Y^2} \end{aligned}$

• Then for both situations,

$\qquad\qquad \begin{aligned} \small X&=\small 4\cos30=2\sqrt3N\\ \small Y&=\small 3+4\sin30=5N\\ \small R&=\small \sqrt{(2\sqrt3)^2+5^2}\\ &=\small \sqrt{37}N\\\\ or\\\\ \small X_1&=\small 4\cos30=2\sqrt3\,N\\ \small Y_1 &=\small 3+(-4\sin30)=1\,N\\ \small R_1&=\small \sqrt{(2\sqrt3)^2+1^2}\\ &=\small \sqrt{13}\,N \end{aligned}$

• Both methods yield the same result & the drawing method should also give the same result when performed.