How do I calculate the normal force on a box on an inclined plane?

[John Comtoi]

Homework Statement

A 20kg box has a static coefficient of friction of 0.28. If a force of 200N is applied at an angle of 45deg, is the box able to move. Explain why once the static friction is overcome, the frictional force decreases and the box moves easier.

Homework Equations

F=m*a
Ff=μFN

The Attempt at a Solution

I don't fully understand this concept. I made a diagram with FN going straight down on the box (perpendicular to the ramp), and Fg going directly down. I know how to calculate the frictional force but I am confused on the solution to the problem. Am i supposed to make a right triangle and use trig to find the hypotenuse and the bottom side?

The right triangle I made consisted of FN=200N and Fg= (9.8m/s*20kg). I am pretty sure the Fg is wrong because the hypotenuse of the right triangle would be smaller than one of the other sides. I feel likes I have the general gyst on how to set up on visualizing it, but I can't execute using the variables at hand

 

[Doc Al]

Realize that the applied force affects the normal force between box and surface. You need to calculate that normal force. Start by identifying all the forces acting on the box in a free body diagram.

 

[John Comtoi]

Realize that the applied force affects the normal force between box and surface. You need to calculate that normal force. Start by identifying all the forces acting on the box in a free body diagram.

https://gyazo.com/373eea5a0624f1562ed194fd5074fd02
https://gyazo.com/373eea5a0624f1562ed194fd5074fd02

Is the normal force not 200N? or am I supposed to factor in the mass of 20kg?

 

[Doc Al]

https://gyazo.com/373eea5a0624f1562ed194fd5074fd02
https://gyazo.com/373eea5a0624f1562ed194fd5074fd02

Is the normal force not 200N? or am I supposed to factor in the mass of 20kg?

Does that diagram relate to this problem? You didn't mention an incline in your first post.

Yes, the weight of the mass will affect the normal force.

 

[Doc Al]

I made a diagram with FN going straight down on the box (perpendicular to the ramp), and Fg going directly down.

I guess you did mention a ramp, but your diagram shows 40 degrees. And you state in post 1 that the applied force is at a 45 degree angle.

Can you restate the problem?

 

[Doc Al]

Also... why in the world is this thread titled "momentum and collision" problem! :)

 

[John Comtoi]

Also... why in the world is this thread titled "momentum and collision" problem! :)

I was initially going to put a different problem but I forgot to change the title!

The diagram angle was a mistake, It was 45deg. Not 40deg.

Is the normal force 200N? or am I mistaken in the way that I'm labeling variables and calculating them?

 

[Doc Al]

I was initially going to put a different problem but I forgot to change the title!

The diagram angle was a mistake, It was 45deg. Not 40deg.

No worries. (Give me a new title and I'll change it.)

Is the normal force 200N?

No. The applied force is 200 N. You also must consider the component of the box's weight normal to the surface.

 

[John Comtoi]

No worries. (Give me a new title and I'll change it.)No. The applied force is 200 N. You also must consider the component of the box's weight normal to the surface.

New Title: Friction problem

Is the normal force the mass of the box*gravity so 20kg(9.8m/s^2)=196N?
and if so, is that the force perpendicular to the ramp in the diagram?

 

[Doc Al]

Is the normal force the mass of the box*gravity so 20kg(9.8m/s^2)=196N?

No, that's the weight of the box. It acts vertically down.

and if so, is that the force perpendicular to the ramp in the diagram?

The 200 N force shown in your diagram is the applied force.

Four forces act on the box, but three have components perpendicular to the ramp. (What are they?) What must be the net force on the box perpendicular to the ramp? Use that to solve for the normal force.