A 1.85kg block is held in equilibrium on an incline of angle θ = 67.7° by a horizontal force, F, applied in the horizontal direction. If the

Resolve the forces perpendicular and parallel to the surface of the incline.  They must separately be balanced.

 

Perpendicular

N =  1.85*9.8*cos(67.7°) + F*sin(67.7°)   ...(1)    N is normal force, gravitational accn = 9.8m/s²

 

Parallel

0.256*N + F*cos(67.7°) = 1.85*9.8*sin(67.7°) ...(2)

 

Use (1) in (2)
0.256*(1.85*9.8*cos(67.7°) + F*sin(67.7°)) + F*cos(67.7°) = 1.85*9.8*sin(67.7°)

 

F*(0.256*sin(67.7°) + cos(67.7°)) = 1.85*9.8*(sin(67.7°) - 0.256*cos(67.7°) )

 

F = 1.85*9.8*(sin(67.7°) - 0.256*cos(67.7°) )/(0.256*sin(67.7°) + cos(67.7°))

 

$${\mathtt{F}} = {\frac{{\mathtt{1.85}}{\mathtt{\,\times\,}}{\mathtt{9.8}}{\mathtt{\,\times\,}}\left(\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{67.7}}^\circ\right)}{\mathtt{\,-\,}}{\mathtt{0.256}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{67.7}}^\circ\right)}\right)}{\left({\mathtt{0.256}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{67.7}}^\circ\right)}{\mathtt{\,\small\textbf+\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{67.7}}^\circ\right)}\right)}} \Rightarrow {\mathtt{F}} = {\mathtt{24.359\: \!321\: \!813\: \!047\: \!781\: \!3}}$$

 

F ≈ 24.4 Newtons