Solve Applied Problems Involving Vectors
-Two forces, $\mathbf { F } _ { \mathbf { 1 } }$ and $\mathbf { F } _ { \mathbf { 2 } }$, of magnitude 60 and 70 pounds, respectively, act on an object. The direction of $\mathbf { F } _ { \mathbf { 1 } }$ is $\mathrm { N } 40 ^ { \circ } \mathrm { E }$ and the direction of $\mathrm { F } _ { 2 }$ is $\mathrm { N } 40 ^ { \circ } \mathrm { W }$. Find the magnitude and the direction angle of the resultant force. Express the direction angle to the nearest tenth of a degree.
A) $\| \mathrm { F } \| = 99.37 ; \quad \theta = 93.7 ^ { \circ }$
B) $| \mathbf { F } | = 92.20 ; \quad \theta = 89.4 ^ { \circ }$
C) $\mid \mathbf { F } \| = 92.20 ; \theta = 80 ^ { \circ }$
D) $| F | = 94.63 ; \theta = 87.2 ^ { \circ }$

Question

Quizplus · Accepted Answer

To solve this, we decompose each force into its horizontal and vertical components and then sum these components to find the resultant force. For $\mathbf { F } _ { \mathbf { 1 } }$, the horizontal component is $60 \cos(40^\circ)$ and the vertical component is $60 \sin(40^\circ)$. For $\mathbf { F } _ { \mathbf { 2 } }$, the horizontal component is $-70 \cos(40^\circ)$ (negative because it's to the west) and the vertical component is $70 \sin(40^\circ)$. Adding the horizontal components and the vertical components separately gives the total horizontal and vertical components of the resultant force. The magnitude of the resultant force is found using the Pythagorean theorem, and the direction angle is found using the arctan of the vertical component over the horizontal component. The calculations match the values given in option A.

Solve Applied Problems Involving Vectors -Two Forces F1\mathbf { F } _ { \mathbf { 1 } }F1​

Solve Applied Problems Involving Vectors
-Two Forces $\mathbf { F } _ { \mathbf { 1 } }$