Question

As shown in the diagram, two forces act on an object. The forces have magnitudes F1 = 5.7 N and F2 = 1.9 N. What third force will cause the object to be in equilibrium?

3.8 N at 108° counterclockwise from
F1->

6.3 N at 162° counterclockwise from
F1->

3.8 N at 162° counterclockwise from
F1->

6.3 N at 108° counterclockwise from
F1->

Answer 1

Second option 6.3 N at 162° counterclockwise from

F1->

Step-by-step explanation:

Observe the attached image. We must calculate the sum of all the forces in the direction x and in the direction y and equal the sum of the forces to 0.

For the address x we have:

`-F_3sin(b) + F_1 = 0`

For the address and we have:

`-F_3cos(b) + F_2 = 0`

The forces
`F_1` and
`F_2` are known

`F_1 = 5.7\ N\\\\F_2 = 1.9\ N`

We have 2 unknowns (
`F_3` and b) and we have 2 equations.

Now we clear
`F_3` from the second equation and introduce it into the first equation.

`F_3 = (F_2)/(cos (b))`

Then

`-(F_2)/(cos (b))sin(b)+F_1 = 0\\\\F_1 = (F_2)/(cos (b))sin(b)\\\\F_1 = F_2tan(b)\\\\tan(b) = (F_1)/(F_2)\\\\tan(b) = (5.7)/(1.9)\\\\tan^(-1)((5.7)/(1.9)) = b\\\\b= 72\°\\\\m = b +90\\\\\m= 162\°`

Then we find the value of
`F_3`

`F_3 = (F_1)/(sin(b))\\\\F_3 =(5.7)/(sin(72\°))\\\\F_3 = 6.01 N`

Finally the answer is 6.3 N at 162° counterclockwise from

F1->