Need help Newtons laws two diemnsions (vertical plane)

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Homework Help Overview

The problem involves analyzing the forces acting on a lawnmower being pushed down at an angle, specifically focusing on Newton's laws in a two-dimensional vertical plane context. The original poster attempts to determine the acceleration of the lawnmower given the applied force, friction, and mass.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of forces in both vertical and horizontal directions, questioning the signs used in the equations. There is an exploration of how to correctly express the normal force and the components of the applied force.

Discussion Status

Participants are actively engaging in clarifying the setup of the problem, with some providing guidance on the importance of correctly identifying forces and their directions. There is a focus on ensuring that the equations reflect the physical situation accurately, particularly regarding the signs of the forces involved.

Contextual Notes

There is an emphasis on the need for careful attention to signs in the equations, as well as the suggestion to draw a free body diagram to visualize the forces acting on the lawnmower. The original poster's understanding of the coefficient of friction is also questioned.

flawlessbrown
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Homework Statement


A gardener pushes down on the handle of a lawnmover, applying a force of 250N. The handle is inclined at an angle 45* to the horizontal. If the coefficient of the kinetic friction between the wheels of the lawnmower and the ground is 0.40, what is the acceleration of the 20-kg lawnmower?

Homework Equations


Fnet=0=Fn+Fsin45-Fg
Fn=mg-FSin45
a=(Fx-Ff)/m
Fnet=ma
Ff=(UkFn) Uk=coefficient of friction miew? or meiu

The Attempt at a Solution



Answer is 1.38m/s^2
 
Last edited:
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Your first equation applies on the y direction, but you have a signage error. The weight acts down and so does the vertical component of the applied force. The normal force acts up ; solve for it. Also, be sure to identify the horizontal component of the applied force, Fx.
 
PhanthomJay said:
Your first equation applies on the y direction, but you have a signage error. The weight acts down and so does the vertical component of the applied force. The normal force acts up ; solve for it. Also, be sure to identify the horizontal component of the applied force, Fx.

so are you saying my expression should be Fn= -mg+Fsin45 ?
 
You must be very careful with plus and minus signs, or else they will get the better of you. . Draw a free body diagram of the mower. The weight acts down on it, the vertical component of the applied force on the handle acts down on it, and the normal force of the ground acts up on it (normal contact forces are generally 'pushing' forces). Choose the up direction as positive and the down direction as negative. The algebraic sum of these three forces in the y direction must equal zero. Watch plus and minus signs, and solve for N.
 

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