force of friction on the cart is given as
[tex]F_f = 280 N[/tex]
here we also know the reaction force due to surface
[tex]R = 766 N[/tex]
so we can say reaction force is given as
[tex]R = \sqrt{F_n^2 + F_f^2}[/tex]
[tex]766 = \sqrt{F_n^2 + 280^2}[/tex]
[tex]F_n^2 = 766^2 - 280^2[/tex]
[tex]F_n = 713 N[/tex]
now by force balance we will say
[tex]F_y + F_n = mg[/tex]
[tex]F_y = mg - F_n[/tex]
[tex]F_x = F_f[/tex]
also we know that
[tex]F_f = \mu * F_n[/tex]
[tex]280 = \mu * 713[/tex]
[tex]\mu = 0.39[/tex]
now minimum force required to set this into motion
[tex]F_{min} = \frac{\mu mg}{\sqrt{1 + \mu^2}}[/tex]
here we know that
[tex]mg = F_n = 713 N[/tex]
[tex]F_{min} = \frac{0.39* 713}{\sqrt{1 + 0.39^2}}[/tex]
[tex]F_{min} = 259 N[/tex]
So it will require 259 N minimum force to move it
