a body is acted on by forces F1=16N,180°,F2= 50 N ,040°,F3=28,150°,calculate the magnitude (correct to one decimal)and the direction(to the nearest degree)of the resultant of the three forces
I don't know the answer help me solve it
$R_x = \Sigma F_x$
$R_x = 16 \cos 180^\circ + 50 \cos 040^\circ + 28 \cos 150^\circ$
$R_y = \Sigma F_y$
$R_x = 16 \sin 180^\circ + 50 \sin 040^\circ + 28 \sin 150^\circ$
$R = \sqrt{{R_x}^2 + {R_y}^2}$
Direction of resultant: $\tan \theta_x = \dfrac{R_y}{R_x}$
Just follow the solution above with your calculator.
More information about text formats
Follow @iMATHalino
MATHalino
I don't know the answer help me solve it
$R_x = \Sigma F_x$
$R_x = 16 \cos 180^\circ + 50 \cos 040^\circ + 28 \cos 150^\circ$
$R_y = \Sigma F_y$
$R_x = 16 \sin 180^\circ + 50 \sin 040^\circ + 28 \sin 150^\circ$
$R = \sqrt{{R_x}^2 + {R_y}^2}$
Direction of resultant:
$\tan \theta_x = \dfrac{R_y}{R_x}$
Just follow the solution above with your calculator.
Add new comment