here we can say that net force on the student vertically upwards will be counter balance by his weight downwards
Let net force F is exerted by each hand
so here we will have
[tex]2Fcos15 = W[/tex]
[tex]2F cos15 = 800[/tex]
[tex]F = \frac{800}{2 cos15}[/tex]
[tex]F = 414.1 N[/tex]
so the force exerted by each hand will be 414.1 N
