Need help Newtons laws two diemnsions (vertical plane)

AI Thread Summary
A gardener applies a force of 250N at a 45-degree angle to push a 20-kg lawnmower, with a kinetic friction coefficient of 0.40. The discussion highlights the importance of correctly applying Newton's laws in two dimensions, particularly focusing on the vertical forces involved. A signage error in the initial equations is pointed out, emphasizing the need to account for the weight and the vertical component of the applied force accurately. Participants suggest drawing a free body diagram to clarify the forces acting on the mower and to ensure proper algebraic treatment of the forces. The final calculated acceleration of the lawnmower is determined to be 1.38 m/s².
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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
 
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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.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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