Question 11.105: A homeowner uses a snowblower to clear his driveway. Knowin...

Vector Mechanics for Engineers: Dynamics-Solution Manual [EXP-4784]

A homeowner uses a snowblower to clear his driveway. Knowing that the snow is discharged at an average angle of 40° with the horizontal, determine the initial velocity ${ \nu }_{ 0 }$ of the snow.

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The snow is discharged at an average angle of 40° with the horizontal.

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Step 1:

First, we need to find the initial horizontal and vertical velocities, denoted as (νx)0 and (νy)0 respectively. We can use the given information that the initial velocity, ν0, makes an angle of 40 degrees with the horizontal. Using trigonometry, we can determine that (νx)0 = ν0 cos 40° and (νy)0 = ν0 sin 40°.

Step 2:

For the horizontal motion, we can use the equation x = x0 + (νx)0 t, where x0 is the initial horizontal position (which is 0 in this case) and t is the time. At point B, the horizontal position is 14 ft. Plugging in the values, we get 14 = (ν0 cos 40°) t. Solving for t, we find tB = 14 / (ν0 cos 40°).

Step 3:

For the vertical motion, we can use the equation y = y0 + (νy)0 t - (1/2) g t^2, where y0 is the initial vertical position (which is 0 in this case) and g is the acceleration due to gravity (32.2 ft/s^2). At point B, the vertical position is 1.5 ft. Plugging in the values and using the value of tB from step 2, we get 1.5 = (ν0 sin 40°) tB - (1/2) g * tB^2.

Step 4:

Now, we can substitute the expression for tB from step 2 into the equation from step 3 to eliminate tB. Simplifying the equation, we obtain ν0^2 = [(1/2) g (14 / (ν0 cos 40°))^2] / (-1.5 + 14 tan 40°).

Step 5:

Finally, we can solve for ν0 by taking the square root of both sides of the equation from step 4. Plugging in the given values for g, 14, and the trigonometric values, we find that ν0 is approximately equal to 22.9 ft/s.

Final Answer

First note

$\begin{aligned}& \left(\nu_x\right)_0=\nu_0 \cos 40^{\circ} \\& \left(\nu_y\right)_0=\nu_0 \sin 40^{\circ}\end{aligned}$

Horizontal motion. (Uniform)

$x=0+\left(\nu_x\right)_0 t$

At B: $\quad \quad \quad 14=\left(\nu_0 \cos 40^{\circ}\right) t \quad \text { or } \quad t_B=\frac{14}{\nu_0 \cos 40^{\circ}}$

Vertical motion. (Uniformly accelerated motion)

$y=0+\left(\nu_y\right)_0 t-\frac{1}{2} g t^2 \quad\left(g=32.2 \ \text{ft} / s ^2\right)$

At B: $\quad \quad \quad 1.5=\left(\nu_0 \sin 40^{\circ}\right) t_B-\frac{1}{2} g t_B^2$

Substituting for $t_B$

$1.5=\left(\nu_0 \sin 40^{\circ}\right)\left\lgroup\frac{14}{\nu_0 \cos 40^{\circ}}\right\rgroup-\frac{1}{2} g\left\lgroup\frac{14}{\nu_0 \cos 40^{\circ}}\right\rgroup^2$

or $\quad \quad \quad \nu_0^2=\frac{\frac{1}{2}(32.2)(196) / \cos ^2 40^{\circ}}{-1.5+14 \tan 40^{\circ}}$

or $\quad \quad \quad \quad \quad \quad \nu_0=22.9 \ \text{ft} / s \blacktriangleleft$

Question 11.105: A homeowner uses a snowblower to clear his driveway. Knowin...

Final Answer

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