Calculating Normal Force on a Box

[PascalPanther]

Susie has a box that weighs 75N that is lying on top of a horizontal floor that is not frictionless. She pushes on the box with a force of 40N directed at an angle 41 degrees below the horizontal. Calculate the normal force that the floor exerts on the box.
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We have not learned anything about friction yet, and it's not in the chapter, so I have no idea if there being friction matters or not for this problem. But ignoring that, it seems easy simple enough.
I break the force exerted into the x and y components.
40*cos41 = 30.1 N
40*sin41 = 26.2 N
x is pointed in the positive direction (+30.1N), and y is pointed in the negative direction (-26.2N). Since the box weighs 75N, it has a force of 75N downward, and that means the normal force is currently 75N upwards. Now, my question is, if everything I did so far is correct, is the new normal force just the total downward force (75N + 26.2N) = 101.2N, or is it equal to the sum of all the forces? Sqrt((75N+26.2N)^2+(30.1N)^2) = 105N?

 

[arildno]

The normal force from the ground must balance out the other forces acting upon the box in the normal direction.
(I.e, the box experience zero acceleration in the normal direction).

 

[berkeman]

You have the right instinct. The total normal force opposes the weight plus the vertical component of the pushing force. Good job. Welcome to PF, BTW.

 

[neutrino]

It's just 75+26.2 N.

 

[neutrino]

Aren't we all excited to see a problem already worked out? ;)

 

[berkeman]

Aren't we all excited to see a problem already worked out? ;)

Yep!

 

[Thunderer]

Okay, so I think I am getting normal force now from that example. So if I apply what is in this problem for my inclined problem, this should be the same I believe.
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A 30000kg truck traveling at 31.3 m/s goes up a 100m frictionless inclined plane. What angle must the inclined plane be above the horizontal for the truck to stop before falling off.

V(i) = 31.3 m/s
V(f) = 0 m/s
Since it is constant, a = 0, and there is no initial force. So normal force cancels out just the mass, and I just need to find the the downward or parallel force.

First thing I did:
Find the acceleration needed to stop in that distance:
V(f)^2 – v(i)^2 = 2a(x(f)-x(i))
31.3 m/s ^2 = 2a(100m)
a =-4.9 m/s^2

I use this equation: a = g * sin O
-4.9m/s^2/-9.8m/s^2 = sin O = O = 30 degrees

Normal force means mass never matters on its own?

 

[cMon89]

Here is a question
A force,F, is acting against the wall on a book of mass,m,. Find the minimum mu so that there is equilibrium.