Question 2.5: Pulling a lawn mower You pull horizontally on the handle of ......
Pulling a lawn mower
You pull horizontally on the handle of a lawn mower that is moving across a horizontal grassy surface. The lawn mower’s mass is 32 kg. You exert a force of magnitude 96 N on the mower. The grassy surface exerts an 83-N resistive friction-like force on the mower. Earth exerts a downward force on the mower of magnitude 314 N, and the grassy surface exerts an upward normal force of magnitude 314 N. What is the acceleration of the mower?
Sketch and translate We choose the mower as the system and sketch the situation, as shown below.
S imp l i f y a n d d i a g r am We model the mower as a point-like object and draw a force diagram for it. As the forces do not change during motion, the diagram can represent the motion at any instant. Three external objects interact with the system—the grassy surface, Earth, and you. You exert a horizontal force on the mower \vec{F}_{\mathrm{Y} \text { on } \mathrm{M}} Earth exerts a downward gravitational force on the mower \vec{F}_{\mathrm{E}} \text { on } \mathrm{M} . The grassy surface exerts a force on the mower that we will represent with two force arrows—a normal force \vec{N}_{\text {Gon M }} perpendicular to the surface (in this case it points upward) and a horizontal friction force \vec{f}_{\mathrm{G} \text { on } \mathrm{M}} opposite the direction of motion.
The downward force \vec{F}_{\text {E on M }} and the upward normal force \vec{N}_{\mathrm{G}} \text { on } \mathrm{M} have the same magnitudes and point in opposite directions. This means that these two forces cancel and do not contribute to the horizontal acceleration of the mower. We can ignore them.
Represent mathematically Since the acceleration of the system is along the x axis and the forces perpendicular to the grassy surface cancel, we can use the x component form of Newton’s second law to determine the acceleration of the mower:
a_{\mathrm{M} x}=\frac{\Sigma F_x}{m_{\mathrm{M}}}=\frac{F_{\mathrm{Y} \text { on } \mathrm{M} x}+f_{\mathrm{G} \text { on } \mathrm{M} x}}{m_{\mathrm{M}}}
The x-component of the force you exert on the mower is positive since that force points in the positive x-direction \left(F_{\mathrm{Y} \text { on } \mathrm{M} x}=+F_{\mathrm{Y} \text { on } \mathrm{M}}\right) . The x-component of the friction force exerted by the grass on the mower is negative since that force points in the negative x-direction \left(f_{\mathrm{G} \text { on } \mathrm{M} x}=-f_{\mathrm{G} \text { on } \mathrm{M}}\right) . Therefore, the acceleration of the mower is
a_{\mathrm{M} x}=\frac{F_{\mathrm{Y} \text { on } \mathrm{M}}+\left(-f_{\mathrm{G} \text { on } \mathrm{M}}\right)}{m_{\mathrm{M}}}=\frac{F_{\mathrm{Y} \text { on } \mathrm{M}} f_{\mathrm{G} \text { on } \mathrm{M}}}{m_{\mathrm{M}}}
Solve and evaluate We can now determine the mower’s acceleration by substituting the known information into the preceding equation:
a_{\mathrm{M} x}=\frac{F_{\mathrm{Y} \text { on } \mathrm{M}}-f_{\mathrm{G} \text { on } \mathrm{M}}}{m_{\mathrm{M}}}=\frac{96 \mathrm{~N}-83 \mathrm{~N}}{32 \mathrm{~kg}}
= +0.40625 N/kg = +0.41m/s²
The known information had two significant digits, so we rounded the answer to two significant digits. The units for acceleration are correct, and the magnitude is reasonable.
Try it yourself: Imagine that you are pulling the mower as described above. After the mower has accelerated to a velocity with which you are comfortable, what is the magnitude of the force you should exert on the mower so that it now moves at constant velocity?