Question #2d7d4

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Answer 1 · 2017-09-20T11:06:46

The first vector $\to V_{1}$ $vec(V_1)$ has a magnitude of 11.7 m and is angled 35.8° counterclockwise from the +x direction.

so $\to V_{1} = 11.7 cos 35.8 ˆ i + 11.7 sin 35.8 ˆ j$ $vec(V_1)=11.7cos35.8hati+11.7sin35.8hatj$

$\Rightarrow \to V_{1} = 9.5 ˆ i + 6.8 ˆ j$ $=>vec(V_1)=9.5hati+6.8hatj$

The 2ndt vector $\to V_{2}$ $vec(V_2)$ has a magnitude of16.5 m and is angled 13.0° counterclockwise from the -x direction.

so $\to V_{2} = 16.5 cos 193 ˆ i + 16.5 sin 193 ˆ j$ $vec(V_2)=16.5cos193hati+16.5sin193hatj$

$\Rightarrow \to V_{2} = - 16.0 ˆ i - 3.7 ˆ j$ $=>vec(V_2)=-16.0hati-3.7hatj$

So resultant vector

$\Rightarrow \to V = \to V_{1} + \to V_{2} = - 7.5 ˆ i + 3.1 ˆ j$ $=>vec(V)=vec(V_1)+vec(V_2)=-7.5hati+3.1hatj$

(a) so magnitude of resultant vector

$\Rightarrow ∣ ∣ ∣ \to V ∣ ∣ ∣ = ∣ ∣ - 7.5 ˆ i + 3.1 ˆ j ∣ ∣ = 8.1 m$ $=>abs(vec(V))=abs(-7.5hati+3.1hatj)=8.1m$

(b) And the resultant makes an angle

${tan}^{- 1} (- \frac{3.1}{7.5}) = {157.5}^{\circ}$ $tan^-1(-3.1/7.5)=157.5^@$ counter clockwise with +x axis