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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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Most popular questions from this chapter

In Fig. 4.15 an astronaut is playing shuffleboard on Earth. The puck has a mass of \(2.0 \mathrm{kg} .\) Between the board and puck the coefficient of static friction is 0.35 and of kinetic friction is \(0.25 .\) (a) If she pushes the puck with a force of \(5.0 \mathrm{N}\) in the forward direction, does the puck move? (b) As she is pushing, she trips and the force in the forward direction suddenly becomes \(7.5 \mathrm{N} .\) Does the puck move? (c) If so, what is the acceleration of the puck along the board if she maintains contact between puck and stick as she regains her footing while pushing steadily with a force of \(6.0 \mathrm{N}\) on the puck? (d) She carries her game to the Moon and again pushes a moving puck with a force of \(6.0 \mathrm{N}\) forward. Is the acceleration of the puck during contact more, the same, or less than on Earth? Explain. (tutorial: rough table)
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