A 10 kg box rest on a table

1. Mar 31, 2015

Nahsor1

1. The problem statement, all variables and given/known data
weight of the box 10 kg on table at rest.
2. Relevant equations
deternmine weight of box and normal reaction acting on it.
also determine w and Fn if force 40N pushes down the box.

3. The attempt at a solutions
i did like
w=mg =NORMAL REACTION(R)
and
40=10 * acceleration due to gravity.
i feeel like this is wrong.

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2. Mar 31, 2015

BvU

Hello Nahsor, welcome to PF !

$W = mg$ seems excellent to me. And if the box doesn't move, the normal force has to be $F_N = - mg$, because then the net force on the box (in a vertical direction) is $F_{\rm net} = F_{\rm gravity} + F_N = mg - mg = 0$.

And your feeling of doubt in the next part is very justified.

Where you find the 10 in your next expression is a mystery to me.
Simply continue the reasoning that if the box doesn't move, the net force has to be 0. Only now htere is an extra force acting. W is related to the box only, so it is and it stays $W = mg$.

3. Mar 31, 2015

Nahsor1

is this the answer for the next part
Fn=40N+10*9.8

4. Mar 31, 2015

BvU

That's correct. You found it from $F_{\rm net} = 0 = -40 \;{\rm N} + mg + F_N$

As you can see, I use the convention up = positive.
And mg is pointing downwards, so if I write $F_{\rm gravity} = mg$ I have to take g = -9.8 m/s2.
Result: $F_N = 40 \;{\rm N} - 10 \;{\rm kg} * (-9.8 \; {\rm m/s^2})$ .

Important tip: always use dimensions -- and check them !

5. Mar 31, 2015

Nahsor1

THANKS FOR URS HELP BVU

6. Mar 31, 2015

Nahsor1

I UNDERSTOOD NOW FORCE UPWARD=FORCE DOWNWARD

7. Mar 31, 2015

BvU

Yes, that's all there's to it.

8. Mar 31, 2015

haruspex

One small correction to the question statement. 10kg is the mass of the box, not the weight. The weight is 10g N.

9. Apr 4, 2015

oh yus yus.