Answer:
(a) See image attached
(b) 486.7 N
(c) 103.06 N
Explanation:
(a)
The free body diagram is attached
(b)
Normal force (N)
[tex]N= mg-Fsin\theta[/tex]
Where m is the mass, g for acceleration due to gravity, F for pulling force and [tex]\theta[/tex] is angle of inclination
Substituting 70 Kg for m, [tex]9.81 m/s^{2}[/tex] for g, 400 N for F and [tex]30^{\circ}[/tex] for [tex]\theta[/tex] we obtain
N=(70*9.81)-(400sin 30)=486.7 N
N=486.7 N
(c)
[tex]F_{net}=Fcos\theta-Fr[/tex] where [tex]F_{net}[/tex] is the net force on horizontal axis, Fr is frictional force which is given by
[tex]Fr=\mu N[/tex]
[tex]F_{net}=Fcos\theta- \mu N[/tex]
Substituting 400 for F, [tex]30^{\circ}[/tex] for [tex]\theta[/tex], 0.5 for [tex]\mu[/tex] and 486.7 for N we obtain
[tex]F_{net}=(400cos30^{\circ})-(0.5*486.7)=103.0601615\approx 103.06 N[/tex]
Therefore, horizontal force is 103.06 N
