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\(\bullet$$\bullet\) A winch is used to drag a 375 \(\mathrm{N}\) crate up a ramp at a con- stant speed of 75 \(\mathrm{cm} / \mathrm{s}\) by means of a rope that pulls parallel to the surface of the ramp. The rope slopes upward at \(33^{\circ}\) above the horizontal, and the coefficient of kinetic friction between the ramp and the crate is 0.25 . (a) What is the tension in the rope? (b) If the rope were suddenly to snap, what would be the accel- eration of the crate immediately after the rope broke?

Short Answer

Expert verified
(a) The tension in the rope is 283 N. (b) The acceleration is 3.3 \text{ m/s}^2.

Step by step solution

01

Calculate the ramp angle

Since the rope is at an angle of \( 33^{\circ} \) to the horizontal, the ramp angle \( \theta \) is also \( 33^{\circ} \).
02

Resolve gravitational force

The component of gravitational force along the ramp is given by \( F_{\text{gravity}} = mg \sin \theta \). Given \( mg = 375 \text{ N} \), the force along the ramp is \( 375 \sin 33^{\circ} \approx 204.3 \text{ N} \).
03

Calculate frictional force

The normal force \( F_{\text{normal}} \) is perpendicular to the ramp. It can be calculated using \( F_{\text{normal}} = mg \cos \theta = 375 \cos 33^{\circ} \approx 314.8 \text{ N} \). The frictional force can be found as \( F_{\text{friction}} = \mu_k F_{\text{normal}} = 0.25 \times 314.8 \approx 78.7 \text{ N} \).
04

Calculate tension in the rope

Since the crate is moved at a constant speed, the net force is zero. This means the tension \( T \) in the rope must balance out the gravitational and frictional forces. Therefore, \( T - F_{\text{gravity}} - F_{\text{friction}} = 0 \), hence \( T = 204.3 + 78.7 = 283 \text{ N} \).
05

Calculate acceleration if rope snaps

If the rope snaps, the only forces acting parallel to the ramp are the gravitational component and friction. Using Newton's second law, \( F_{\text{net}} = ma \), where \( a \) is acceleration. \( F_{\text{net}} = F_{\text{gravity}} - F_{\text{friction}} \). Thus, \( a = (204.3-78.7)/ (375/9.8) \approx 3.3 \text{ m/s}^2 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Newton's Second Law
Newton's Second Law of Motion is a fundamental principle in physics, represented by the equation \( F = ma \), where \( F \) is the net force acting on an object, \( m \) is its mass, and \( a \) is its acceleration. In the context of the crate problem, Newton's second law helps us understand how forces act on the crate and how they affect its motion. Since the crate is moving at a constant speed, the net force acting on it is zero. This is because a constant speed indicates that no acceleration is present, meaning the forces pulling and resisting the crate are balanced.

When the rope pulls the crate up the ramp, the critical forces include the gravitational force pulling it down the ramp and the frictional force resisting its motion. By ensuring that the sum of these forces equals the tension in the rope, Newton's second law supports finding the force needed to maintain the constant speed. If the rope breaks, Newton's second law also helps to determine the crate's acceleration due to the imbalance of forces, where only gravity and friction now act parallel to the ramp.
Kinetic Friction
Kinetic friction is the force that opposes the movement of two surfaces sliding past one another. In physics problems like this one involving a crate on a ramp, kinetic friction plays a crucial role. This frictional force is calculated using the formula \( F_{\text{friction}} = \mu_k F_{\text{normal}} \), where \( \mu_k \) is the coefficient of kinetic friction and \( F_{\text{normal}} \) is the normal force.

The normal force in this scenario is the component of the gravitational force acting perpendicular to the ramp. It is determined by \( F_{\text{normal}} = mg \cos \theta \). For our crate, with a weight of 375 N, this calculation gives a normal force of approximately 314.8 N when the ramp angle is 33 degrees. The coefficient of kinetic friction \( \mu_k \) is given as 0.25, so the frictional force opposing the crate's movement is about 78.7 N.

Kinetic friction must be considered when calculating the net forces involved and plays a key part in determining both the tension in the rope and any resulting acceleration if the rope were to snap.
Rope Tension
Rope tension in physics refers to the pulling force transmitted through a rope, string, or cable when it is pulled tight by forces acting at each end. This tension must be sufficient to overcome any opposing forces, such as gravity and friction, to move an object, like our crate, at a desired constant speed.

In the problem, the tension in the rope must balance out both the component of gravitational force pulling the crate back down the ramp and the kinetic friction opposing its motion. Since the crate moves at constant speed, the net force is zero, meaning \( T = F_{\text{gravity}} + F_{\text{friction}} \).

The tension calculation involves adding the gravitational component along the ramp, approximately 204.3 N, to the frictional force opposing movement, about 78.7 N, resulting in a total tension of 283 N. Accurately calculating rope tension is crucial for understanding how much force is required to ensure the crate moves without accelerating, reflecting the direct application of Newton's laws in practical scenarios.

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Most popular questions from this chapter

\(\bullet$$\bullet\) You've attached a bungee cord to a wagon and are using it to pull your little sister while you take her for a jaunt. The bungee's unstretched length is \(1.3 \mathrm{m},\) and you happen to know that your little sister weighs 220 \(\mathrm{N}\) and the wagon weighs 75 \(\mathrm{N}\) . Crossing a street, you accelerate from rest to your normal walk- ing speed of 1.5 \(\mathrm{m} / \mathrm{s}\) in 2.0 \(\mathrm{s}\) , and you notice that while you're accelerating, the bungee's length increases to about 2.0 \(\mathrm{m}\) . What's the force constant of the bungee cord, assuming it obeys Hooke's law?

\(\bullet$$\bullet\) Friction in an elevator. You are riding in an elevator on the way to the 18 th floor of your dormitory. The elevator is accelerating upward with \(a=1.90 \mathrm{m} / \mathrm{s}^{2} .\) Beside you is the box containing your new computer; the box and its contents have a total mass of 28.0 \(\mathrm{kg} .\) While the elevator is accelerating upward, you push horizontally on the box to slide it at constant speed toward the elevator door. If the coefficient of kinetic friction between the box and the elevator floor is \(\mu_{\mathrm{k}}=0.32\) what magnitude of force must you apply?

What is the acceleration of a raindrop that has reached half of its terminal velocity? Give your answer in terms of \(g .\)

the task of designing an accelerometer to be used inside a rocket ship in outer space. Your equipment consists of a very light spring that is \(15.0 \mathrm{~cm}\) long when no forces act to stretch or compress it, \(\begin{array}{lllll}\text { plus } & \text { a } & 1.10 & \text { kg } & \text { weight. }\end{array}\) be attached to a friction- free tabletop, while the \(1.10 \mathrm{~kg}\) weight is attached to the other end, as shown in Figure 5.67 . (Such a spring-type accelerometer system was actually used in the ill-fated Genesis Mission, which collected particles of the solar wind. Unfortunately, because it was installed backward, it did not measure the acceleration correctly during the craft's descent to earth. As a result, the parachute failed to open and the capsule crashed on Sept. \(8,2004 .)\) (a) What should be the force constant of the spring so that it will stretch by \(1.10 \mathrm{~cm}\) when the rocket accelerates forward at \(2.50 \mathrm{~m} / \mathrm{s}^{2}\) ? Start with a free-body diagram of the weight. (b) What is the acceleration (magnitude and direction) of the rocket if the spring is compressed by \(2.30 \mathrm{~cm} ?\)

\(\bullet$$\bullet\) Atwood's Machine. A 15.0 -kg load of bricks hangs from one end of a rope that passes over a small, frictionless pulley. A 28.0 -kg counterweight is suspended from the other end of the rope, as shown in Fig. 5.60 . The system is released from rest. (a) Draw two free-body diagrams, one for the load of bricks and one for the counterweight. (b) What is the magnitude of the upward acceleration of the load of bricks? (c) What is the ten- sion in the rope while the load is moving? How does the ten- sion compare to the weight of the load of bricks? To the weight of the counterweight?

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