Question

Find the angle θ between forces F and P.

Express your answer to three significant figures in degrees.

Answer 1

We need to find the vector P in order to find the angle between F and P

So, Given magnitude of P = 40 N

Direction angles $\alpha =132^{\circ}$ $\beta =87.0^{\circ}$$\gamma =42.2^{\circ}$

We know a unit vector is given by

$\hat{P}=cos\alpha\, \hat{i} +cos\beta \, \hat{j}+cos\gamma \, \hat{k}$

$\Rightarrow \vec{P}=\left | P \right |(cos\alpha\, \hat{i} +cos\beta \, \hat{j}+cos\gamma \, \hat{k})$

$\Rightarrow \vec{P}=40(cos132^{\circ}\, \hat{i} +cos87^{\circ} \, \hat{j}+cos42.2^{\circ} \, \hat{k})$

$\Rightarrow \vec{P}=40(-0.7\, \hat{i} +0.05 \, \hat{j}+0.74\, \hat{k})$

$\Rightarrow \vec{P}=-28\, \hat{i} +2 \, \hat{j}+29.6\, \hat{k} N$

To find the angle we need to find the dot product as

$\vec{a}.\vec{b}=\left |\vec{a} \right |\left |\vec{b} \right |cos\theta$, where $\theta$ is the angle between the two vectors.

$\Rightarrow cos\theta=\frac{\vec{a}.\vec{b}}{\left |\vec{a} \right |\left |\vec{b} \right |}$

Here,

$\vec{F}= 8.80\, \hat{i}-10.5\, \hat{j}+22.8\, \hat{k}$

$\left |\vec{F} \right |= \sqrt{8.8^{2}+(-10.5)^{2}+22.8^{2}}=26.6N$

If $\theta$ is the angle between F and P, then

$\Rightarrow cos\theta=\frac{\vec{F}.\vec{P}}{\left |\vec{F} \right |\left |\vec{P} \right |}$

$\Rightarrow cos\theta=\frac{( 8.80\, \hat{i}-10.5\, \hat{j}+22.8\, \hat{k}).(-28\, \hat{i} +2 \, \hat{j}+29.6\, \hat{k})}{26.6 *40}$

$\Rightarrow cos\theta=\frac{-246.4-21+674.88}{1064}$

$\Rightarrow cos\theta=\frac{407.48}{1064}$

$\Rightarrow cos\theta=0.382$

$\Rightarrow \theta=\cos^{-1}0.382$

$\therefore \theta=67.542^{\circ}$

which is the required answer.

If any doubt feel free to comment.