Question

Find the net force acting on a box that has a force of 50 newtons pulling at an angle of 30 degrees and a force of 90 newtons pulling at a 160 degree angle​

Answer 1

Approximately
`69\; {\rm N}` at approximately
`126^(\circ)`.

Step-by-step explanation:

Assume that both angles in the question are relative to the positive
`x`-axis (towards the positive horizontal direction.)

Horizontal component (
`x`-component) of the two forces would be:

• 
  `(50\; {\rm N}) \, \cos(30^(\circ)) = \left(25\, √(3)\right)\; {\rm N} \approx 43.3\; {\rm N}`.
• 
  `(90\; {\rm N}) \, \cos(160^(\circ)) \approx (-84.6) \; {\rm N}`.

Note that the
`x`-component of the
`90\; {\rm N}` force is negative since this components points away from the positive
`x\!`-direction.

Hence, the net force in the
`x`-component would be:

`(50\; {\rm N}) \, \cos(30^(\circ)) + (90\; {\rm N}) \, \cos(160^(\circ)) \approx (-41.3) \; {\rm N}`.

(Again, this component is negative since it points away from the positive
`x`-axis.)

Similarly, the vertical component (
`y`-component) of the two forces would be:

• 
  `(50\; {\rm N}) \, \sin(30^(\circ)) = 25\; {\rm N}`.
• 
  `(90\; {\rm N}) \, \sin(160^(\circ)) \approx 30.8 \; {\rm N}`.

Hence, the net force in the
`y`-component would be:

`(50\; {\rm N}) \, \sin(30^(\circ)) + (90\; {\rm N}) \, \sin(160^(\circ)) \approx 55.8\; {\rm N}`.

Refer to the diagram attached. The resultant net force is the vector sum of the components. Apply the Pythagorean Theorem to find this net force:

`\begin{aligned}(\text{net force}) &= \sqrt{(\text{$x$-component})^(2) + (\text{$y$-component})^(2)} \\ &\approx \sqrt{(-41.3)^(2) + (55.8)^(2))}\; {\rm N} \\ &\approx 69\; {\rm N}\end{aligned}`.

Find the angle of this net force relative to the positive
`x`-axis using the inverse cosine function
`\arccos`:

`\begin{aligned}\arccos\left(\frac{(\text{$y$-component})}{(\text{net force})}\right) &\approx \arccos\left(\frac{55.8\; {\rm N}}{69\; {\rm N}}\right) \\ &\approx 126^(\circ)\end{aligned}`.

(The units might need to be converted into degrees.)