Two tugboats pull a barge across the harbor. One boat exerts a force of
7.5 × 104 N north, while the second boat exerts a force of 9.5 × 104 N at
15.0° north of west. Precisely, in what direction does the barge move?
To determine the direction in which the barge moves, we need to find the net force acting on the barge by combining the forces exerted by the two tugboats. We can do this by adding the individual force vectors.
The first tugboat exerts a force of 7.5 × 10^4 N north. Since this force is only in the north direction, we can represent it as a vector with a magnitude of 7.5 × 10^4 N and a direction of 0° (north).
The second tugboat exerts a force of 9.5 × 10^4 N at 15.0° north of west. To represent this vector, we need to break it down into its horizontal and vertical components. The horizontal component can be found using the cosine function, and the vertical component can be found using the sine function.
Horizontal Component = 9.5 × 10^4 N * cos(15°)
Vertical Component = 9.5 × 10^4 N * sin(15°)
Now, to find the net force, we can add the horizontal components and the vertical components separately.
Horizontal Net Force = 0 - (9.5 × 10^4 N * cos(15°))
Vertical Net Force = 7.5 × 10^4 N + (9.5 × 10^4 N * sin(15°))
To calculate the magnitudes, we add the squared values of the horizontal and vertical components and take the square root of the sum.
Net Force Magnitude = √(Horizontal Net Force^2 + Vertical Net Force^2)
Finally, to find the direction in which the barge moves, we need to find the angle formed by the net force vector with respect to the positive x-axis. This angle can be found using the arctan function.
Direction of the Barge = arctan(Vertical Net Force / Horizontal Net Force)
By plugging in the values and calculating, we can determine the precise direction in which the barge moves.