College

A car traveling on a [tex]$2.7^{\circ}$[/tex] uphill grade has a grade resistance of 112 lb.

What is the approximate weight of the car in pounds?

[tex]$\square$[/tex] lb.

Answer :

We are given that the component of the car's weight acting parallel to the slope (the grade resistance) is 112 lb when the car is on an uphill grade of [tex]$2.7^\circ$[/tex]. This component is given by

[tex]$$
\text{grade resistance} = W \sin \theta,
$$[/tex]

where [tex]$W$[/tex] is the weight of the car and [tex]$\theta$[/tex] is the angle of the grade.

Step 1. Write down the formula relating the grade resistance and the weight:

[tex]$$
W \sin \theta = 112 \text{ lb}.
$$[/tex]

Step 2. Substitute the angle [tex]$\theta = 2.7^\circ$[/tex] into the formula:

[tex]$$
W \sin (2.7^\circ) = 112.
$$[/tex]

Step 3. Solve for [tex]$W$[/tex] by dividing both sides of the equation by [tex]$\sin(2.7^\circ)$[/tex]:

[tex]$$
W = \frac{112}{\sin(2.7^\circ)}.
$$[/tex]

Step 4. Compute the sine of [tex]$2.7^\circ$[/tex]. When computed (ensuring that the angle is in degrees), we obtain

[tex]$$
\sin(2.7^\circ) \approx 0.04712389.
$$[/tex]

Step 5. Substitute the value of [tex]$\sin(2.7^\circ)$[/tex] into the equation for [tex]$W$[/tex]:

[tex]$$
W \approx \frac{112}{0.04712389} \approx 2377.59 \text{ lb}.
$$[/tex]

Thus, the weight of the car is approximately [tex]$2377.59$[/tex] lb.