Newton's second and third law problem

[APysics]

Homework Statement

A horizontal force(with respect to the ground) is exerted on a 2 kg box on an incline. theta is 60 degrees above the horizontal. what force Fa is needed to keep the box stationary on the incline? what force does the incline exert on the box?

Homework Equations

F=ma

The Attempt at a Solution

netfx=M1a
Fax-19.6sin60=0
Fax=19.6sin60

netfy=m1a
Fn-Fay+19.6cos60=0
Fn= -19.6cos60+Fay

Im not sure even if I started it correctly, need help!

Homework Statement

Homework Equations

The Attempt at a Solution

 

[collinsmark]

netfx=M1a

Okay, that's right. That's Newton's second law.

Fax-19.6sin60=0
Fax=19.6sin60

But that's not right.

Your first step is to draw a block diagram. If you haven't done so already, do so now.

You should notice (after drawing your block diagram) that there are only two force components in the x-direction. There is the F_a, which is completely in the x-direction because the problem statement says it is a horizontal force; and there is the x-component of the normal force.

netfy=m1a
Fn-Fay+19.6cos60=0
Fn= -19.6cos60+Fay

There isn't an F_ay term (or more accurately, I should say F_ay = 0). The problem statement says, "A horizontal force(with respect to the ground)..."

Also, take another look at your block diagram. Is the y-component of the normal force equal to mg, or is the y-component of mg equal to the normal force?

 

[APysics]

okay so after that i get:
netfx=m1a
Fa-19.6sin60=m1a

netfy=m1a
fn-19.6cos60=0
Fn=19.6cos60

and I am stuck again, is this right so far?

 

[collinsmark]

okay so after that i get:
netfx=m1a
Fa-19.6sin60=m1a

Sorry, still not quite right. The magnitude of the Normal force is not 19.6 N.

netfy=m1a
fn-19.6cos60=0
Fn=19.6cos60

According to the above equation, you are expressing the normal force as a component of mg. But that's not right. Look at your block diagram again. mg is equal to a component of the normal force. Not the other way around.

 

[rwisz]

First, let's define some variables:
- Fa: the horizontal force applied to the box
- Fn: the normal force exerted by the incline on the box
- Fg: the force of gravity on the box
- m: mass of the box (2 kg in this case)
- g: acceleration due to gravity (9.8 m/s^2)

Using Newton's second law, we can set up two equations:
1) In the x-direction:
Fa - Fg*sin(theta) = 0 (since the box is not moving horizontally, the net force in the x-direction is 0)
Solving for Fa, we get:
Fa = Fg*sin(theta)

2) In the y-direction:
Fn - Fg*cos(theta) = 0 (since the box is not moving vertically, the net force in the y-direction is 0)
Solving for Fn, we get:
Fn = Fg*cos(theta)

Now, substituting in the values for Fg and theta, we get:
Fa = (2 kg)*(9.8 m/s^2)*sin(60 degrees) = 19.6 N
Fn = (2 kg)*(9.8 m/s^2)*cos(60 degrees) = 9.8 N

Therefore, the force needed to keep the box stationary on the incline is 19.6 N, and the force exerted by the incline on the box is 9.8 N. This makes sense because the incline is supporting some of the weight of the box, reducing the normal force needed to keep it stationary.