The resultant of the concurrent forces is 300 lb pointing up along the Y axis. Compute the values of F and the angle required to

Question

xenn [34]

4 years ago

12

The resultant of the concurrent forces is 300 lb pointing up along the Y axis. Compute the values of F and the angle required to

give the resultant.

Physics

2 answers:

dsp73 · Answer 1 · 2022-01-06T20:09:42+03:00

dsp734 years ago

5 0

<span>500 cos(180) i + 500 sin(180)
240 cos(-30) i + 240 sin(-30)
F cos(theta) i + F sin(theta)
-----------------------------------
300 cos(90) i + 300 sin(90)</span>

Nikolay [14] · Answer 2 · 2022-01-06T22:24:06+03:00

The values of F and the angle required to give the resultant are F = 823.529 lbs and theta = 30.667 degrees

<h3>Further explanation </h3>

The resultant of the concurrent forces is 300 lb pointing up along the Y axis. Compute the values of F and the angle required to give the resultant.

The first thing is to get the x and y components of each forces, then isolate the given figure:

The sum of y components is

$\sum F_y = R_y\\F sin \theta - (240 lbs) (sin 30) = 300 lbs\\F sin \theta - (240 lbs) (0.5) = 300 lbs\\F sin \theta - (120 lbs) = 300 lbs\\F sin \theta = 420 lbs$

The sum of x components is

$\sum F_x = R_x\\F cos \theta - (240 lbs) (cos 30) - 500 lbs = 0\\F cos \theta - (240 lbs) (0.866) - 500 lbs = 0\\F cos \theta - (207.846 lbs) - 500 lbs = 0\\F cos \theta = 707.846 lbs$

Substitute the value of θ to either of the two working equations, we have

$\frac{F sin \theta}{F cos \theta} = \frac{420 lbs}{707.846 lbs} \\tan \theta = 0.593\\\theta = 30.667 \\ F sin \theta = 420 lbs\\F sin 30.667 = 420 lbs\\F = \frac{420 lbs}{sin 30.667} \\F = \frac{420 lbs}{0.510} \\F = 823.529 lbs$