The resultant of two forces P and Q is 1200 N horizontally leftward. Determine the force Q and corresponding angle ? for the system of force s as shown in Fig. 1a.

Given,

$R_x = -1200\ N$

$R_y = 0$

Now,

$\sum F_x = R_x$

$\implies -Q\ sin\theta + 600\ cos\ 60 = -1200$

$\implies Q\ sin\ theta = 600\ cos\ 60 + 1200$

$\implies \therefore Q\ sin \theta = 1500 ......(i)$

Since, $R_y = 0$

$\implies -Q\ cos\theta + 60\ sin\ 60 = 0$

$\implies Q\ cos\ \theta = 60\ sin\ 60$

$\implies \therefore Q\ cos\theta = 519.615 ........(ii)$

Dividing equation (i) by (ii), we have -

$\implies \frac{Q\ sin\theta}{Q\ cos\theta} = \frac{1500}{519.615} = tan \theta$

$\implies \therefore\ \theta = 70.89^o Ans.$