Answer:
[tex]F=185.006N[/tex]
Explanation:
Summing moments about the wheelbarrow-to-ramp contact point to zero
∑[tex]F=F_y*L-F_N[/tex]
[tex]F_y[L]-29.35kg*(9.81)*[\frac{L}{2}]=0[/tex]
[tex]Fy=\frac{m*g}{2}=\frac{29.35kg*9.8m/s^2}{2}=143.8N[/tex]
Summing forces in the vertical direction to zero
[tex]F_N*cos(36) + Fy - 29.35kg(9.8m/s^2)=0[/tex]
[tex]F_N*cos(39)= 29.35kg(9.8m/s^2)-143.8N[/tex]
[tex]F_N=185.07N[/tex]
Summing horizontal forces to zero
[tex]F_N*sin(39)- F_x=0[/tex]
[tex]F_x=116.47N[/tex]
total force
[tex]F=\sqrt{F_x^2+F_y^2}=\sqrt{116.47^2+143.8^2}[/tex]
[tex]F=185.006N[/tex]
which is interestingly at an angle of 36.0° from the vertical.
