Two crewmen pull a boat through a lock. One crewman pulls with a force of 130 N at an angle of 35o relative to the forward direction of the boat. The second crewman, on the opposite side of the lock, pulls at an angle for 45o. With what force should the second crewman pull so that the net force of the two crewmen is in the forward direction?

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Because the boatmen are pulling at angles relative to the boat, the force with which they are pulling can be split up into two component vectors, with the rope acting like the hypotenuse of a right triangle formed by the component vectors. Remember that any straight-line vector can be described as the product of two vectors at right angles to each other.

The first boatman is pulling at an angle of 35 degrees. This means that some of his 130N force is devoted to pulling the boat through the lock, and some is devoted to pulling the boat toward himself. We can use trigonometry to determine what these two forces are.

The sine of an angle describes the ratio of the opposite side, divided by the hypotenuse. This would be the force component that moves the boat toward the boatman. This will be some amount less than 130N, so we multiply 130N by the sine of 35

130 sin 35 = 74.56N

On the other side of the boat, the other boatman is doing the same thing. However, his angle is different. What needs to take place here, in order for the boat to move only in the forward direction, is for the second boatman to pull with enough force that exactly 74.56N of force are pulling the boat toward himself. This way, the two forces cancel out, and only the forward forces remain.

The second boatman pulls at an angle of 45 degrees. The sine of this angle, times the second boatman's force, must equal 74.56N.

F cos 45 = 74.56

F = `(74.56)/(cos 45)`

**F = 105.44N**

Fcos45>= 130cos35

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