Pushing a Lawnmower: Find tanθ_critical

  • Thread starter phthiriasis
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In summary, the lawnmower of weight w can be pushed at a constant speed by applying a force F_{h}, parallel to the handle at an angle theta. The magnitude of this force can be determined using the equation F_{h}= (-\mu w)/(-cos\theta+sin\theta\mu). However, this force becomes infinite when the denominator becomes 0, which occurs at an angle \theta_{critical} where cos\theta_{critical} = \mu sin\theta_{critical}. Therefore, the expression for tan\theta_{critical} is 1/\mu.
  • #1
phthiriasis
2
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Homework Statement


Consider a lawnmower of weight w which can slide across a horizontal surface with a coefficient of friction [tex]\mu[/tex]. In this problem the lawnmower is pushed using a massless handle, which makes an angle theta with the horizontal. Assume that [tex]F_{h}[/tex], the force exerted by the handle, is parallel to the handle.

Take the positive x direction to be to the right and the postive y direction to be upward.

http://img408.imageshack.us/img408/2788/mfscf8ayr4.jpg
http://g.imageshack.us/img408/mfscf8ayr4.jpg/1/

A:
Find the magnitude, [tex]F_{h}[/tex] of the force required to slide the lawnmower over the ground at constant speed by pushing the handle.

B:


The solution for [tex]F_{h}[/tex] has a singularity (that is, becomes infinitely large) at a certain angle [tex]\theta_{critical}[/tex]. For any angle [tex]\theta[/tex] > [tex]\theta_{critical}[/tex], the expression for [tex]F_{h}[/tex] will be negative. However, a negative applied force [tex]F_{h}[/tex] would reverse the direction of friction acting on the lawnmower, and thus this is not a physically acceptable solution. In fact, the increased normal force at these large angles makes the force of friction too large to move the lawnmower at all.

Find an expression for tan[tex]\theta_{critical}[/tex]


2. The attempt at a solution

okay, so i got part A fine:


fnet=0
x:
-cos[tex]\theta[/tex]*[tex]F_{h}[/tex]+[tex]F_{f}[/tex]
y:
n-sin[tex]\theta[/tex]*[tex]F_{h}[/tex]-w

multiply the y equation by [tex]\mu[/tex] to eliminate n
-cos[tex]\theta[/tex]*[tex]F_{h}[/tex]+[tex]\mu[/tex]n=[tex]\mu[/tex]n-sin[tex]\theta[/tex][tex]F_{h}[/tex][tex]\mu[/tex]

rearrange to get fh...
[tex]F_{h}[/tex]= (-[tex]\mu [/tex]w)/(-cos[tex]\theta[/tex]+sin[tex]\theta[/tex][tex]\mu[/tex])

part b i just not sure where to start..the force becomes infinite when the denominator becomes 0... but i don't know how to express that in terms of tan[tex]\theta_{critical}[/tex].

thanks for any help
lol i see now someone had the exact same question.. no answer though..here
 
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  • #2
Welcome to PF!

phthiriasis said:
[tex]F_{h}[/tex]= (-[tex]\mu [/tex]w)/(-cos[tex]\theta[/tex]+sin[tex]\theta[/tex][tex]\mu[/tex]

part b i just not sure where to start..the force becomes infinite when the denominator becomes 0... but i don't know how to express that in terms of tan[tex]\theta_{critical}[/tex].

Hi phthiriasis! Welcome to PF! :smile:

(have a theta: θ and a mu: µ :smile:)

Yes, you're right …

the force becomes infinite when the denominator becomes 0 …

ie cosθcritical = µsinθcritical

so tanθcritical = … ? :smile:
 
  • #3
1/µ
seems so simple now... thanks tim
 

1. What is tanθ_critical?

Tanθ_critical is the critical angle of incline at which a lawnmower will start to tip over while being pushed.

2. How is tanθ_critical calculated?

Tanθ_critical is calculated using the formula tanθ_critical = μ_s / μ_k, where μ_s is the coefficient of static friction and μ_k is the coefficient of kinetic friction.

3. What factors affect tanθ_critical?

The coefficient of friction, the weight and distribution of weight of the lawnmower, and the surface on which the lawnmower is being pushed all affect tanθ_critical.

4. Why is it important to know tanθ_critical?

Knowing tanθ_critical helps determine the maximum angle of incline at which a lawnmower can safely be pushed without tipping over. This can prevent accidents and injuries.

5. Can tanθ_critical be different for different lawnmowers?

Yes, tanθ_critical can vary depending on the weight, size, and distribution of weight of the lawnmower. It can also vary based on the surface on which the lawnmower is being pushed.

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