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Chapter 4: Problem 10

A barge is hauled along a straight-line section of canal by two horses harnessed to tow ropes and walking along the tow paths on either side of the canal. Each horse pulls with a force of \(560 \mathrm{N}\) at an angle of \(15^{\circ}\) with the centerline of the canal. Find the sum of the two forces exerted by the horses on the barge.

Short Answer

Expert verified
Answer: The total force exerted by the horses on the barge is approximately \(1080.66 \mathrm{N}\), acting along the centerline of the canal.

Step by step solution

01

Define the given forces as vectors

The given forces can be defined as vectors with magnitudes and directions. Let \(\textbf{F}_1\) and \(\textbf{F}_2\) be the forces exerted by the two horses. The magnitudes of both forces are \(560 \mathrm{N}\), and they both form an angle of \(15^\circ\) with the centerline of the canal.
02

Find the horizontal and vertical components of each force

We can use trigonometry to find the horizontal (x-axis) and vertical (y-axis) components of each force. The horizontal component of \(\textbf{F}_1\) is given by \(F_{1x} = F_1\cos (\theta)\), and the vertical component of \(\textbf{F}_1\) is given by \(F_{1y} = F_1\sin (\theta)\). Similarly, we can find the components of \(\textbf{F}_2\). Here, \(\theta = 15^\circ\). \(F_{1x} = 560\cos (15^\circ)\) \(F_{1y} = 560\sin (15^\circ)\) \(F_{2x} = 560\cos (15^\circ)\) \(F_{2y} = -560\sin (15^\circ)\) (The vertical component of \(\textbf{F}_2\) is negative since it acts in the opposite direction)
03

Add the horizontal and vertical components of the forces

Now that we have the horizontal and vertical components of each force, we can add them to find the total force exerted by the horses on the barge. \(F_{total_x} = F_{1x} + F_{2x} = 2 \times 560\cos (15^\circ)\) \(F_{total_y} = F_{1y} + F_{2y} = 560\sin (15^\circ) - 560\sin (15^\circ) = 0\) The total force has a horizontal component of \(2 \times 560\cos (15^\circ)\) and a vertical component of \(0\).
04

Find the magnitude and direction of the total force

Since the vertical component of the total force is 0, the total force is entirely horizontal and acts along the centerline of the canal. The magnitude of the total force is equal to the horizontal component: \(F_{total} = 2 \times 560\cos (15^\circ)\) \(F_{total} ≈ 1080.66 \mathrm{N}\) The total force exerted by the horses on the barge is approximately \(1080.66 \mathrm{N}\), acting along the centerline of the canal.

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