Magnitude of Normal Force (Frictionless)

[Spartan Erik]

Homework Statement

"A crate of mass 50 kg is pushed across a frictionless horizontal floor with a force of 100N directed 23.5 degrees below the horizontal. The magnitude of the normal force of the floor on the crate is:"

Homework Equations

F = ma? Not sure what else would apply.. mainly a conceptual issue here

The Attempt at a Solution

I took 100N x cos(23.5) in order to get the horizontal force which turns out to by 91.7N.. not sure what to do from here

 

[Doc Al]

To solve for the normal force, analyze the vertical components of the forces acting on the crate. (Three forces act.)

 

[Spartan Erik]

Well gravity is acting on the crate (9.8 m/s^2 downward), and the crate itself has a normal force exerted in the opposite direction

 

[Doc Al]

Gravity and the normal force are two of the three forces. What's the third vertical component?

 

[Spartan Erik]

Well I imagine the third component could be sin(23.5 degrees) x 100 = 39.875

 

[Doc Al]

Well I imagine the third component could be sin(23.5 degrees) x 100 = 39.875

Right. So what must the normal force be to balance this force plus gravity?

 

[Spartan Erik]

Ah so 9.8 m/s^2 x 50kg = 490N

And adding that to 39.875N will result in a magnitude of 530N