Newton's second and third law problem

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Homework Help Overview

The problem involves analyzing the forces acting on a 2 kg box on an incline with a 60-degree angle. The original poster seeks to determine the horizontal force required to keep the box stationary and the normal force exerted by the incline.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to apply Newton's second law to find the forces in both x and y directions. Some participants question the accuracy of the force components and suggest drawing a block diagram to clarify the forces involved.

Discussion Status

Participants are actively discussing the setup of the problem and the application of Newton's laws. There is ongoing clarification regarding the components of forces, particularly the normal force and its relationship to the weight of the box. No consensus has been reached yet, and participants are exploring different interpretations of the problem.

Contextual Notes

There is uncertainty regarding the correct expression of the normal force and its components, as well as the role of the horizontal force in the context of the incline. The original poster expresses doubt about their initial approach, indicating a need for further guidance.

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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

 
Last edited:
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APysics said:
netfx=M1a
Okay, that's right. That's Newton's second law. :approve:
Fax-19.6sin60=0
Fax=19.6sin60
But that's not right. :frown:

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 Fa, 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 Fay term (or more accurately, I should say Fay = 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?
 
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?
 
APysics said:
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.
 

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