Answer:
Fh = u*m*g / (cos(θ) - u*sin(θ))
Explanation:
Given:
- The mass of lawnmower = m
- The angle the handle makes with the horizontal = θ
- The force applied along the handle = Fh
- The coefficient of friction of the lawnmower with ground = u
Find:
Find the magnitude, Fh, of the force required to slide the lawnmower over the ground at constant speed by pushing the handle.
Solution:
- Construct a Free Body Diagram (FBD) for the lawnmower.
- Realize that there is horizontal force applied parallel to ground due to Fh that drives the lawnmower and a friction force that opposes this motion. We will use to Newton's law of motion to express these two forces in x-direction as follows:
F_net,x = m*a
- Since, the lawnmower is to move with constant speed then we have a = 0.
F_net,x = 0
- The forces as follows:
Fh*cos(θ) - Ff = 0
Where, Ff is the frictional force:
Fh = Ff /cos(θ)
Similarly, for vertical direction y the forces are in equilibrium. Using equilibrium equation in y direction we have:
- W - Fh*sin(θ) + Fn = 0
Where, W is the weight of the lawnmower and Fn is the contact force exerted by the ground on the lawnmower. Then we have:
Fn = W + Fh*sin(θ)
Fn = m*g + Fh*sin(θ)
The Frictional force Ff is proportional to the contact force Fn by:
Ff = u*Fn
Ff = u*(m*g + Fh*sin(θ))
Substitute this expression in the form derived for Fh and Ff:
Fh*cos(θ) = u*(m*g + Fh*sin(θ))
Fh*(cos(θ) - u*sin(θ)) = u*m*g
Fh = u*m*g / (cos(θ) - u*sin(θ))
