Respuesta :
ou should have found that
F
h
FhF_h
, the force required to push the lawnmower at constant speed, was
F
h
=
μmg
cos(θ)−μsin(θ)
Fh=μmgcos(θ)−μsin(θ)
.
Note that this expression becomes infinite when the denominator equals zero:
cos(θ)−μsin(θ)=0
cos(θ)−μsin(θ)=0
,
or
tan(
θ
critical
)=
1
μ
tan(θcritical)=1μ
.
(The phrase "
F
h
FhF_h
has a singularity at angle
θ
critical
θcriticaltheta_critical
" means that "
F
h
FhF_h
goes to infinity at a certain angle
θ
critical
θcriticaltheta_critical
.")
It's not too hard to understand what this means. Suppose you were pushing straight down on the lawnmower (
θ=90
θ=90
degrees). It obviously wouldn't move. But, according to the equation for
F
h
FhF_h
, when you plug in
θ=90
θ=90
degrees, you get a negative force (which doesn't make sense).
The more vertical you push, the harder it gets to move the lawnmower. At
θ=
θ
critical
θ=θcritical
, it gets impossible to move it. The force required to move it goes to infinity; you have to push infinitely hard.
Provide Feedback
Next
Correct. Followup.
You should have found that , the force required to push the lawnmower at constant speed, was
.
Note that this expression becomes infinite when the denominator equals zero:
,
or
.
(The phrase " has a singularity at angle " means that " goes to infinity at a certain angle .")
It's not too hard to understand what this means. Suppose you were pushing straight down on the lawnmower ( degrees). It obviously wouldn't move. But, according to the equation for , when you plug in degrees, you get a negative force (which doesn't make sense).
The more vertical you push, the harder it gets to move the lawnmower. At , it gets impossible to move it. The force required to move it goes to infinity; you have to push infinitely hard.
End of followup.
The force need to slide the lawnmower over the ground with constant velocity by pushing the handle is,
[tex]F_h=\dfrac{\mu mg}{\cos \theta-\mu \sin\theta}[/tex]
The expression for tan (θcritical),
[tex]\tan(\theta_{crtical})=\dfrac{1}{\mu}[/tex]
What is force?
Force is the effect of pull or push due to which the object having a mass changes its velocity.
The force is of two types-
- Push-When the force applied in the direction of motion of the object, then the force is called the push force.
- Pull- When the force applied in the opposite direction of motion of the object, then the force is called the pull force.
It is given that the mass of the lawnmower is m.
Coefficient of friction of horizontal surface is μ.
The angle made between the handle and horizontal is θ.
The force exerted by the handle, which is parallel to the handle is [tex]F_h[/tex].
Forces acting on the lawnmower are-
- The normal force acting on it is,
[tex]F_N=F_h\cos \theta[/tex]
- The friction force applied on it is given as,
[tex]F_f=\mu F_N\\F_f=\mu(mg+ F_h\sin\theta)[/tex]
- The applied horizontal force actin is given as,
[tex]F=F_h\cos \theta[/tex]
The force required to slide the lawnmower over the ground at constant speed by pushing the handle is,
On solving the above equation we get,
[tex]F_h=\dfrac{\mu mg}{\cos \theta-\mu \sin\theta}[/tex]
Expression for tan(θcritical)-
The above equation becomes the infinite, when the value of denominator is equal to zero as,
[tex]\cos \theta-\mu\sin \theta=0[/tex]
Solve it further,
[tex]\cos \theta=\mu\sin \theta\\\dfrac{\sin \theta}{\cos \theta}=\dfrac{1}{\mu}[/tex]
Ratio of sine and cosine is tangent theta. Thus,
[tex]\tan(\theta_{crtical})=\dfrac{1}{\mu}[/tex]
Hence, the force required to slide the lawnmower over the ground at constant speed by pushing the handle is,
[tex]F_h=\dfrac{\mu mg}{\cos \theta-\mu \sin\theta}[/tex]
The expression for tan(θcritical)-
[tex]\tan(\theta_{crtical})=\dfrac{1}{\mu}[/tex]
Learn more about the force here;
https://brainly.com/question/388851
