Question

Three forces act on an object. Two of the forces have the magnitudes 56 N and 27 N, and make angles 53° and 156°, respectively, with the positive x-axis. Find the magnitude (in N) and the direction angle (in degrees) from the positive x-axis of the third force such that the resultant force acting on the object is zero. (Round your answers to two decimal places. Let the direction angle be greater than 0° and less than 360°.)

Answer 1

Step-by-step explanation:

Given

Two forces
`F_1` and
`F_2` at an angle of
`\theta _1`
`\theta _2`

`F_1=56\ N`

`F_2=27\ N`

`\theta _1=53^(\circ)`

`\theta _2=156^(\circ)`

As resultant force is zero therefore horizontal component as well as vertical component of force is zero

`\sum F_x=F_1\cos \theta _1+F_2\cos \theta _2+F_3\cos \theta _3`

`\sum F_x=56\cos 53+27\cos 156+F_3\cos \theta`

`F_3\cos \theta_3=-9.035\ N----1`

`\sum F_y=F_1\sin \theta _1+F_2\sin \theta _2+F_3\sin \theta _3`

`F_3\sin \theta _3=55.704\ N----2`

squaring and adding 1 and 2

`F_3^2(\cos ^2\theta _3+\sin^2\theta _3)=86.583+3102.93`

`F_3=56.47\ N`

Divide 1 and 2 to get
`\theta _3`

`(\sin \theta_3 )/(\cos \theta_3 )=(-55.704)/(9.305)`

`\tan \theta_3=-6.16`

`\theta _3=180-80.78`

`\theta _3=99.22^(\circ)`