1. Mar 30, 2005

xangel31x

1)A gardener pushes a 12 kg lawnmower whose handle is tilted up 37 degrees above horizontal. The lawnmower's coefficient of rolling friction is 0.15. How much power does the gardener have to supply to push the lawnmower at a constant speed of 1.2 m/s?

ok so i have tried this problem and still can't seem to get it right..

i am taking power to be equal to the force*velocity so i have been doing this

mass*gravity*cos(37)*friction*velocity=power...i have no idea what else to try. i need some suggestions...

i have tried finding mass*gravity*cos(37) and subtracting if from mass*gravity*friction and then multiplying that number by the velocity and that was wrong to.

2. Mar 30, 2005

xanthym

From the problem statement:
{Mower Mass} = M = (12 kg)
{Handle Angle} = θ = (37 deg)
{Coeff Friction} = μ = (0.15)
{Speed} = v = (1.2 m/s)
{Gardener's Force on Mower} = F

Since the mower is moving at constant speed, the horizontal component of the gardener's force must exactly balance the horizontal friction force:
{Friction Force} = F*cos(θ)

The friction force will be derived from the combination of mower's weight and gardener's vertical force component:
{Friction Force} = μ*(M*g + F*sin(θ))

Equating the 2 above equations:
F*cos(θ) = μ*(M*g + F*sin(θ))
::: ⇒ F*cos(θ) = μ*M*g + μ*F*sin(θ)
::: ⇒ F*(cos(θ) - μ*sin(θ)) = μ*M*g
::: ⇒ F = μ*M*g/(cos(θ) - μ*sin(θ))
::: ⇒ F = M*g/(cos(θ)/μ - sin(θ))
::: ⇒ {Power} = F*v = v*M*g/(cos(θ)/μ - sin(θ))
::: ⇒ {Power} = (1.2 m/s)*(12 kg)*(9.81 m/sec^2)/(cos(37 deg)/(0.15) - sin(37 deg))
::: ⇒ {Power} = (29.91 Watts)

~~~

Last edited: Mar 30, 2005
3. Mar 30, 2005

xangel31x

thank you for the help