Chapter 4, Problem 48P

A person pushes a 14.0-kg lawn mower at constant speed with a force of F = 88.0 N directed along the handle, which is at an angle of 45.0° to the horizontal (Fig. 4-58) (a) Draw the free-body diagram showing all forces acting on the mower. Calculate (b) the horizontal friction force on the mower, then (c) the normal force exerted vertically upward on the mower by the ground (d) What force must the person exert on the lawn mower to accelerate it from rest to 1.5 m/s in 2.5 seconds, assuming the same friction force?

Figure 4-58 Problem 50.

To determine

Part (a) To Determine:

A free body diagram for the motion.:

Answer to Problem 48P

Solution:

The free body diagram for the lawn mower is drawn as below:

Explanation of Solution

Given:

Mass of the lawn mower $m=14.0\text{ kg}$

Force acting on the mower $F=88.0\text{ N}$

The angle at which the force acts $\theta =45.0°$

The initial speed of the mower $v_0=0\text{ m/s}$

Final speed of the mower $v=1.5\text{ m/s}$

Time taken to reach the final speed $t=2.5\text{ m/s}$

Calculation:

The free body diagram for the lawn mower is given below:

The lawn mower of mass m is acted upon by a force F at an angle $\theta$. The force is resolved into two components, parallel to the ground and perpendicular to the ground. The weight mg of the mower acts vertically downwards and the normal force $F_N$ acts vertically upwards. The force of friction $F_{fr}$ acts along the ground in a direction opposite to the motion of the mower

To determine

Part (b)To determine:

The horizontal frictional force on the lawn mower

Answer to Problem 48P

Solution:

The frictional force when the lawn mover moves with constant speed is 62.2 N.

Explanation of Solution

Given:

Mass of the lawn mower $m=14.0\text{ kg}$

Force acting on the mower $F=88.0\text{ N}$

The angle at which the force acts $\theta =45.0°$

The initial speed of the mower $v_0=0\text{ m/s}$

Final speed of the mower $v=1.5\text{ m/s}$

Time taken to reach the final speed $t=2.5\text{ m/s}$

Calculation:

The lawnmower moves with a constant speed under the action of the force F. Therefore, the net force on the mower is zero.

From the free body diagram, write the condition for the equilibrium in the horizontal direction.

    $F_{fr}=F\cos \theta$

Substitute the values of F and $\theta$

from the given values and calculate the value of the frictional force.

    $\begin{array}{c}{F}_{fr}=F\cos \theta \end{array}$

        $\begin{array}{c}=\left(88.0\text{ N}\right)\left(\cos 45°\right)\end{array}$

        $\begin{array}{c}=62.2\text{ N}\end{array}$

To determine

Part (c)To determine:

The normal force acting on the mower

Answer to Problem 48P

Solution:

The normal force acting on the mower is199 N.

Explanation of Solution

Given:

Mass of the lawn mower $m=14.0\text{ kg}$

Force acting on the mower $F=88.0\text{ N}$

The angle at which the force acts $\theta =45.0°$

The initial speed of the mower $v_0=0\text{ m/s}$

Final speed of the mower $v=1.5\text{ m/s}$

Time taken to reach the final speed $t=2.5\text{ m/s}$

Calculation:

Since there is no motion, either in the upward or in the downward directions, write the conditions for equilibrium in the vertical direction.

From the free body diagram,

    $F_N=mg+F\sin \theta$

Use the values for m, F and $\theta$

from the given values and using 9.8m/s² for g in the equation, calculate the value of the normal force.

    $\begin{array}{c}{F}_N=mg+F\sin \theta \end{array}$

        $\begin{array}{c}=\left(14.0\text{ kg}\right)\left(9.8{\text{ m s}}^{-2}\right)+\left(88.0\text{ N}\right)\left(\sin 45°\right)\end{array}$

        $\begin{array}{c}=199.4\text{ N}\\ =199\text{ N}\end{array}$

To determine

Part (d) To Determine:

The force needed to accelerate the lawn mower from rest to 1.5 m/s in 2.5 s when the same frictional force acts on the lawnmower.

Answer to Problem 48P

Solution:

The force required to accelerate the lawn mover from rest to 1.5 m/s in 2.5 s when the same frictional force acts on the lawn mower is 99.6 N.

Explanation of Solution

Given:

Mass of the lawn mower $m=14.0\text{ kg}$

Force acting on the mower $F=88.0\text{ N}$

The angle at which the force acts $\theta =45.0°$

The initial speed of the mower $v_0=0\text{ m/s}$

Final speed of the mower $v=1.5\text{ m/s}$

Time taken to reach the final speed $t=2.5\text{ m/s}$

Calculation:

If the magnitude of the force changes to F₁ and the lawn mower accelerates with an acceleration a, from rest to a speed v in a time t, with the frictional force having the same magnitude as in the previous case, the free body diagram, in this case, is drawn as is shown below:

In the horizontal direction, the equation for motion can be written as,

    $F_1\cos \theta -F_{fr}=ma$

Calculate the acceleration using the equation of motion,

    $a=\frac{v-v_0}{t}$

Substitute the values of $v_0$, v and t and calculate a.

    $\begin{array}{c}a=\frac{v-v_0}{t}\end{array}$

        $\begin{array}{c}=\frac{\left(1.5\text{ m/s}\right)-\left(0\text{ m/s}\right)}{\left(2.5\text{ s}\right)}\end{array}$

        $\begin{array}{c}=0.6{\text{ m s}}^{-2}\end{array}$

Use the calculated values of a and $F_{fr}$ in the equation $F_1\cos \theta -F_{fr}=ma$ and calculate the value of $F_1$.

    $\begin{array}{c}{F}_1\cos \theta =F_{fr}+ma\end{array}$

    $\begin{array}{c}{F}_1\cos 45°=\left(62.2\text{ N}\right)+\left(14.0\text{ kg}\right)\left(0.6{\text{ m s}}^{-2}\right)\end{array}$

        $\begin{array}{c}{F}_1=70.4\sqrt{2}\text{ N}\end{array}$

        $\begin{array}{c}=99.6\text{ N}\end{array}$