Explanations & Calculations
What is asked is the vector addition of 2 vectors and the included angle is given. In such a case, using the equation R 2 = P 2 + Q 2 + 2 P Q cos θ \small R^2=P^2+Q^2+2PQ\cos\theta R 2 = P 2 + Q 2 + 2 PQ cos θ , the magnitude of their resultant (R) can be calculated.
Now the two forces are P ⃗ = 2 A ⃗ \small \vec{P}=2\vec{A} P = 2 A and Q ⃗ = 3 B ⃗ \small \vec{Q}=3\vec{B} Q = 3 B Then the corresponding magnitudes are ∣ P ⃗ ∣ = P = ∣ 2 A ⃗ ∣ = 2 ∣ A ⃗ ∣ = 2 × 4 u n i t s = 8 u n i t s \qquad\small |\vec{P}|=P=|2\vec{A}|=2|\vec{A}|=2\times4units=8\,units ∣ P ∣ = P = ∣2 A ∣ = 2∣ A ∣ = 2 × 4 u ni t s = 8 u ni t s
∣ Q ⃗ ∣ = Q = ∣ 3 Q ⃗ ∣ = 3 ∣ Q ⃗ ∣ = 3 × 6 u n i t s = 18 u n i t s . \qquad\small |\vec{Q}|=Q=|3\vec{Q}|=3|\vec{Q}|=3\times6\,units=18\,units. ∣ Q ∣ = Q = ∣3 Q ∣ = 3∣ Q ∣ = 3 × 6 u ni t s = 18 u ni t s .
Then plugging these data in the above equation, R 2 = 8 2 + 1 8 2 + 2 ( 8 ) ( 18 ) cos 60 = 532 u n i t s 2 R = 23.065 N \qquad\qquad
\begin{aligned}
\small R^2&=\small8^2+18^2+2(8)(18)\cos60\\
&=\small 532\,units^2 \\
\small R &=\small \bold{23.065N}
\end{aligned} R 2 R = 8 2 + 1 8 2 + 2 ( 8 ) ( 18 ) cos 60 = 532 u ni t s 2 = 23.065N
Comments