Question

Braking Load A Toyota Sienna with a gross weight of $5300$pounds is parked on a street with an $8^{\circ }$grade. See the figure. Find the magnitude of the force required to keep the Sienna from rolling down the hill. What is the magnitude of the force perpendicular to the hill?

Short answer

The magnitude of force required to keep the Sienna from rolling down the hill is $736.6$ pounds. The magnitude of the force perpendicular to the hill is$5248.4$pounds.

Step-by-step solution

Step 1. Given Information 

The gross weight of the Toyota Sienna is $5300$pounds. It is parked on a street with $8^{\circ }$grade. We have to find the magnitude of the force required to keep the Sienna from rolling down the hill and the magnitude of the force perpendicular to the hill.

Step 2. Finding the angle

Let angle $B=8^{\circ }$ and angle A and C as adjacent angles but angles Aand B form the alternate angles, thus they are equal.

So, $A=B=8^{\circ }$

$$\begin{gathered} C={90}^{\circ }-A \\ C={90}^{\circ }-8^{\circ } \\ C={82}^{\circ } \end{gathered}$$

The angle between the slope and the weight of the car is${82}^{\circ }$

Step 3. Finding the force

Let was the weight of the car.

Now, wcan be represented as a vector

So,

$$\begin{gathered} \mathbf{w}=5300\left(\cos {82}^{\circ }\mathbf{i}+\sin {82}^{\circ }\mathbf{j}\right) \\ \mathbf{w}=5300\left(0.13917\mathbf{i}+0.99026\mathbf{j}\right) \\ \mathbf{w}=737.6\mathbf{i}\mathbf{}+5248.4\mathbf{j} \end{gathered}$$

In any vector $c=a\mathbf{i}+b\mathbf{j}$, a and b are horizontal and vertical components.

Therefore, the magnitude of force required to keep the sienna from rolling down the hill is $737.6$pounds and the magnitude of the force perpendicular to the hill is$5248.4$pounds.