1. Ben is pulling on a tobaggan rope with a force of 250 newtons. The rope makes a 36 angle with the ground. What force is actually working to move the tobaggan to the right?

2. If a boat going 6 mi/hr in still water suddenly encounters a crosscurrent of 4 mi/hr, at what angle will the boat veer?

Thank you so much!

When pulling on an object with a force at some angle, the upward component is
sin(thetz)*force and the horizontal component is
cos(theta)*force

For the boat let's assume a simple position: it's going east and the crosscurrent is going north. The boat's velocity is
(6mi/hr,0) and the current's is (0,4mi/hr)
If the boat travels for one hour it will travel along some direction a distance 6mi and it will be 4mi north of where it started. Draw a right triangle with hypotenuse 6 and one leg =4 to correspond to the river's velocity. You should get sin(theta)=4/6. Do you see how to set this up now?

1. To determine the force actually working to move the toboggan to the right, we need to determine the horizontal component of the force exerted by Ben.

The force exerted by Ben is 250 newtons, and the angle it makes with the ground is 36 degrees.

To find the horizontal component, we can use the formula: horizontal component = force × cosine(angle)

Plugging in the values, we get: horizontal component = 250 N × cos(36°)

Using a calculator, we find that the cosine of 36 degrees is approximately 0.809.

Therefore, the horizontal component of the force exerted by Ben is: 250 N × 0.809 ≈ 202.25 N.

So, the force actually working to move the toboggan to the right is approximately 202.25 Newtons.

2. To determine the angle at which the boat will veer when encountering the crosscurrent, we can use trigonometry.

Let's assume the boat is traveling east (along the x-axis) at a speed of 6 mi/hr, and encounters a crosscurrent flowing north (along the y-axis) at a speed of 4 mi/hr.

Since the boat will move both horizontally and vertically, we can create a right triangle with the boat's velocity as the hypotenuse.

The horizontal leg of the triangle represents the boat's eastward velocity, which is 6 mi/hr. The vertical leg represents the crosscurrent's northward velocity, which is 4 mi/hr.

To find the angle at which the boat will veer, we can use the tangent function: tangent(theta) = vertical leg / horizontal leg

Plugging in the values, we get: tangent(theta) = 4 mi/hr / 6 mi/hr

Dividing both sides by 4/6, we get: tangent(theta) = 2/3

To find the angle theta, we can take the inverse tangent (arctan) of both sides.

Using a calculator, we find that the angle theta is approximately 33.69 degrees.

Therefore, the boat will veer at an angle of approximately 33.69 degrees.