Questions: Suppose that a boat weighs 700 pounds and is on a ramp inclined at 30 degrees. Represent the force due to gravity, F, using F=-700 j as shown in the figure to the right. a. Write a unit vector along the ramp in the upward direction. b. Find the vector projection of F onto the unit vector from part a. c. What is the magnitude of the vector projection in part b? What does this represent? a. The unit vector along the ramp in the upward direction is u= (Simplify your answer. Type your answer in terms of i and j. Use integers or fractions for any numbers in the expression. Rationalize all denominators. Type an exact answer, using radicals as needed.)

Solution Steps

The ramp is inclined at 30 degrees. The unit vector along the ramp in the upward direction can be represented as: \[ \mathbf{u} = \cos(30^\circ) \mathbf{i} + \sin(30^\circ) \mathbf{j} \]

Using the trigonometric values: \[ \cos(30^\circ) = \frac{\sqrt{3}}{2} \] \[ \sin(30^\circ) = \frac{1}{2} \]

Thus, the unit vector is: \[ \mathbf{u} = \frac{\sqrt{3}}{2} \mathbf{i} + \frac{1}{2} \mathbf{j} \]

The force due to gravity is given by: \[ \mathbf{F} = -700 \mathbf{j} \]

The vector projection of \(\mathbf{F}\) onto \(\mathbf{u}\) is given by: \[ \text{proj}_{\mathbf{u}} \mathbf{F} = \left( \frac{\mathbf{F} \cdot \mathbf{u}}{\mathbf{u} \cdot \mathbf{u}} \right) \mathbf{u} \]

First, calculate the dot product \(\mathbf{F} \cdot \mathbf{u}\): \[ \mathbf{F} \cdot \mathbf{u} = (-700 \mathbf{j}) \cdot \left( \frac{\sqrt{3}}{2} \mathbf{i} + \frac{1}{2} \mathbf{j} \right) = -700 \cdot \frac{1}{2} = -350 \]

Next, calculate the dot product \(\mathbf{u} \cdot \mathbf{u}\): \[ \mathbf{u} \cdot \mathbf{u} = \left( \frac{\sqrt{3}}{2} \mathbf{i} + \frac{1}{2} \mathbf{j} \right) \cdot \left( \frac{\sqrt{3}}{2} \mathbf{i} + \frac{1}{2} \mathbf{j} \right) = \left( \frac{\sqrt{3}}{2} \right)^2 + \left( \frac{1}{2} \right)^2 = \frac{3}{4} + \frac{1}{4} = 1 \]

Thus, the vector projection is: \[ \text{proj}_{\mathbf{u}} \mathbf{F} = (-350) \mathbf{u} = -350 \left( \frac{\sqrt{3}}{2} \mathbf{i} + \frac{1}{2} \mathbf{j} \right) = -175 \sqrt{3} \mathbf{i} - 175 \mathbf{j} \]

The magnitude of the vector projection is given by: \[ \left| \text{proj}_{\mathbf{u}} \mathbf{F} \right| = \sqrt{(-175 \sqrt{3})^2 + (-175)^2} \]

Calculate the squares: \[ (-175 \sqrt{3})^2 = 175^2 \cdot 3 = 91875 \] \[ (-175)^2 = 30625 \]

Add the squares: \[ 91875 + 30625 = 122500 \]

Take the square root: \[ \sqrt{122500} = 350 \]

The magnitude of the vector projection is 350 pounds. This represents the component of the gravitational force acting along the direction of the ramp.

Final Answer