Normal force on 14.8kg box pulled upward by rope

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rachellehowe
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Homework Statement


A 14.8kg box is placed on a table, and a rope, rigged to a pulley, is attached to the top of the box. Using the rope, 3.5 Newtons of force, in the upward direction (ie the rope is pulling the box up), is applied to the box.
a)what would be the normal force exerted on the box by the table?

b) if and 80 kg person sat on the box, and there was no longer a force being exerpted by the rope, what would the new normal force on the box be?


Homework Equations





The Attempt at a Solution



I tried for part a). but I'm not sure if I did it right.
F(normal)= mass*gravity
so i just multiplied 14.8kg by 9.8 m/s/s and got 145.04 Newtons then subtracted the amount from the rope and got 145.04-3.5=141.54Newtons but i don't think its right.. i need help!
 

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rachellehowe said:
I tried for part a). but I'm not sure if I did it right.
F(normal)= mass*gravity
so i just multiplied 14.8kg by 9.8 m/s/s and got 145.04 Newtons then subtracted the amount from the rope and got 145.04-3.5=141.54Newtons but i don't think its right.. i need help!
Sounds good to me (assuming I understand the setup).

Here's a better way to think of this problem. Account for all the forces on the box:
(1) the weight, mg, acting down
(2) the rope tension, 3.5 N, acting up
(3) the normal force from the table, Fn, acting up

These forces must add to zero, since the box is in equilibrium:
Fn + 3.5 - mg = 0

So Fn = mg - 3.5