problem 1.3 - the resultant of the two forces acting on the block is known to be a horizontal 80 - n force…

Explanation:

Step1: Resolve forces horizontally

Let's consider the horizontal - component of the forces. The horizontal component of the 50 - N force is $50\cos25^{\circ}$, and the horizontal component of force $F$ is $F\cos\theta$. The resultant horizontal force $R = 80$ N. So, $F\cos\theta+50\cos25^{\circ}=80$. $F\cos\theta=80 - 50\cos25^{\circ}$ $F\cos\theta=80-50\times0.9063 = 80 - 45.315=34.685$

Step2: Resolve forces vertically

The vertical component of the 50 - N force is $50\sin25^{\circ}$, and the vertical component of force $F$ is $F\sin\theta$. Since the resultant force is horizontal, the sum of the vertical components of the two forces is zero. So, $F\sin\theta-50\sin25^{\circ}=0$. $F\sin\theta = 50\sin25^{\circ}$ $F\sin\theta=50\times0.4226 = 21.13$

Step3: Find the angle $\theta$

Divide the vertical - component equation by the horizontal - component equation: $\tan\theta=\frac{F\sin\theta}{F\cos\theta}=\frac{50\sin25^{\circ}}{80 - 50\cos25^{\circ}}$. $\tan\theta=\frac{21.13}{34.685}\approx0.6092$. $\theta=\arctan(0.6092)\approx31.3^{\circ}$

Step4: Find the magnitude of force $F$

Substitute $\theta$ into the vertical - component equation $F\sin\theta = 50\sin25^{\circ}$. $F=\frac{50\sin25^{\circ}}{\sin\theta}$. Since $\theta\approx31.3^{\circ}$ and $50\sin25^{\circ}=21.13$, $F=\frac{21.13}{\sin31.3^{\circ}}$. $F=\frac{21.13}{0.521}\approx40.6$ N

Answer:

$\theta\approx31.3^{\circ}$, $F\approx40.6$ N