A lawn mower weighs 1274 N. A person tries to push the mower up a hill that is inclined 15o
to the
horizontal. How much force does the person have to exert along the handle just to keep the mower from
rolling down the hill if the normal force exerted on the lawn mower by the ground is 1230.5 N, assuming
that the handle is parallel to the ground?
normalforceonground:
weight*cos15+forcepushing*sin15=1230.5
solve for force pushing
To find the force that the person has to exert along the handle to keep the mower from rolling down the hill, we need to resolve the weight and the normal force into their components along the inclined plane.
First, let's find the weight component that acts parallel to the inclined plane. We can use the formula:
Weight component parallel to the inclined plane = Weight * sin(θ)
where θ is the angle of the hill, which is 15 degrees in this case.
Weight component parallel to the inclined plane = 1274 N * sin(15o)
Weight component parallel to the inclined plane ≈ 330.39 N
Now, to keep the mower from rolling down the hill, the person needs to exert a force equal in magnitude but opposite in direction to the weight component parallel to the inclined plane. Therefore, the force that the person needs to exert along the handle is approximately 330.39 N.
Note: The normal force (1230.5 N) is perpendicular to the inclined plane and does not affect the force required to keep the mower from rolling down the hill in this case.