Answer:
[tex]F_{palms}=392N[/tex]
[tex]F_{feet}=274N}[/tex]
Explanation:
To solve this problem we need to be aware that the man is rotational and mechanical equilibrium, it means that the sum of all torques for a fixed rotation axis is zero and the net force on him is zero too. Since we are working in a two-dimensional space we can work with the magnitude of the torque and considering upwards as positive.
[tex]\tau=r*R*sin(\theta)[/tex],
where F is the force applied, [tex]r*sin(\theta)[/tex] is the perpendicular distance from the axis of rotation to the line of action of the force. In this particular case, all forces are perpendicular to this distance so our expression reduces to
[tex]\tau=r*R[/tex].
Now we can start solvig our problem. We are going to use
Using the feet as the axis of rotation
[tex]\sum \tau=\tau_{feet}+\tau_{palms}-\tau_{weight}=0[/tex],
[tex]\sum \tau=0*F_{feet}+1.7+F_{palms}-1*mg=0[/tex],
[tex]\tau=0*F_{feet}+1.7+F_{palms}-1*mg=0[/tex],
solving for [tex]F_{palms}[/tex]
[tex]F_{palms}=\frac{mg}{1.7}=\frac{68*9.8}{1.7}[/tex]
[tex]F_{palms}=392N[/tex].
To find the force exerted on his feet we use Newton's second law
[tex]\sum F=ma[/tex],
since we are in mechanical equilibrium
[tex]\sum F=0[/tex],
so
[tex]F_{palms}+F_{feet}-mg=0[/tex]
[tex]F_{feet}=mg-F_{palms}[/tex]
[tex]F_{feet}=274N}[/tex].
