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Chapter 13: Problem 90

The \(40-\mathrm{kg}\) boy is sliding down the smooth spiral slide such that \(z=-2 \mathrm{~m} / \mathrm{s}\) and his speed is \(2 \mathrm{~m} / \mathrm{s}\). Determine the \(r, \theta, z\) components of force the slide exerts on him at this instant. Neglect the size of the boy. Prob. \(13-90\)

Short Answer

Expert verified
The components of force the slide exerts on the boy are \(F_r = \frac {160}{r}\) Newton, \(F_胃 = 0N\) and \(F_z = -408.4N \).

Step by step solution

01

Analyze the Due Force

Considering the directions, we can represent the forces as follows: \(F_r = m* a_r\), \(F_胃 = m* a_胃\) and \(F_z = m* a_z - W\). Since the slide is frictionless, there would be no force in the \(胃\) direction (or tangential component). Thus, \(F_胃 = 0\). The acceleration in the \(r\) or radial direction (centripetal acceleration, \(a_r\)) can be represented as \(v^2/r\). Here, \(v\) is the total speed of the boy, and \(r\) is the radius of the slide. Finally, the vertical force \(F_z\) along \(z\) direction is the sum of the mass multiplied by the acceleration due to gravity \(g\) and the vertical acceleration \(a_z\).
02

Derive the Components of Force

Substituting the known values, we get: \(F_r = m * v^2 / r = 40* (2)^2 / r = 160 / r \) N. Since \(F_胃 = 0 N\), and \(F_z = m*a_z - W = 40*(-2) - 40*9.81 = -40.81N\).
03

Submission of the Results

The components of force on the boy due to the slide are \(F_r = 160 / r\) N and \(F_z = -408.4\) N. The negative sign for \(F_z\) indicates that the vertical force is directed downward. The \(F_胃\) component is \(0 N\) as per the assumption.

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Key Concepts

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

Centripetal Force
Centripetal force is a critical concept in understanding motion along curved paths. Imagine a boy sliding down a spiral slide where the path curves in circular form. As he moves, he experiences a force that keeps pulling him towards the center of the curvature. This force is known as the centripetal force.

It's essential because it prevents the boy from flying off the slide and maintains his circular motion. The magnitude of the centripetal force is calculated using the formula: \[ F_r = \frac{m v^2}{r} \]where:
  • \(F_r\) is the radial or centripetal force,
  • \(m\) is the mass of the object (in this case, the boy),
  • \(v\) is the velocity (or speed) of the object,
  • \(r\) is the radius of the circular path.
The centripetal force acts perpendicular to the direction of motion, pulling the boy inward and ensuring he follows the slide鈥檚 curve.
Newton's Second Law
Newton's Second Law of Motion forms the backbone of understanding forces and motion. According to this law, the force acting on an object is equal to the mass of that object times its acceleration: \( F = m \times a \).

In our slide problem, we assess each direction separately using this principle.

Here's how it applies:
  • **Radial Force**: The force in the radial direction is linked to centripetal acceleration, ensuring the boy's circular path. Using Newton鈥檚 law, this force is \( F_r = m \times a_r \).
  • **Tangential Force**: For frictionless slides, there is an absence of tangential or \( F_胃 \) force, simplifying the problem. This is because no force is required to keep him moving in the tangential direction.
  • **Vertical Force**: In the vertical \(z\) direction, we calculate the net force as \( F_z = m \times a_z - W \), where \( W \) (weight) is the force due to gravity. The negative sign in the solution indicates that gravity acts opposite to the boy's motion.
By breaking down the forces in different directions, Newton鈥檚 Second Law helps us understand how they separately contribute to the boy鈥檚 motion.
Motion in Three Dimensions
Understanding motion in three dimensions helps in comprehending how objects move through space. When the boy slides down the spiral slide, his movement can be dissected into multiple components.
  • **Radial Component (\( r \))**: This accounts for the centripetal motion as the boy curves along the slide.
  • **Tangential Component (\( 胃 \))**: In scenarios like this, with a frictionless surface, the tangential component is negligible. No force component is required to either speed up or slow down their motion along the slide.
  • **Vertical Component (\( z \))**: Here, the boy is moving downwards as described by the change in his vertical position. Combining forces due to gravity and vertical acceleration, the net force in the vertical direction (\( F_z \)) is calculated.
Comprehending these three dimensions allows one to visualize and solve complex mechanical problems effectively, providing clarity in how forces act in different spatial directions.

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Most popular questions from this chapter

The \(2-\mathrm{kg}\) pendulum bob moves in the vertical plane with a velocity of \(8 \mathrm{~m} / \mathrm{s}\) when \(\theta=0^{\circ} .\) Determine the initial tension in the cord and also at the instant the bob reaches \(\theta=30^{\circ} .\) Neglect the size of the bob.

A smooth can \(C\), having a mass of \(3 \mathrm{~kg}\), is lifted from a feed at \(A\) to a ramp at \(B\) by a rotating rod. If the rod maintains a constant angular velocity of \(\dot{\theta}=0.5 \mathrm{rad} / \mathrm{s}\) determine the force which the rod exerts on the can at the instant \(\theta=30^{\circ} .\) Neglect the effects of friction in the calculation and the size of the can so that \(r=(1.2 \cos \theta) \mathrm{m}\). The ramp from \(A\) to \(B\) is circular, having a radius of \(600 \mathrm{~mm}\). Prob. 13-95

The 10 -lb block has a speed of \(4 \mathrm{ft} / \mathrm{s}\) when the force of \(F=\left(8 t^{2}\right) \mathrm{lb}\) is applied. Determine the velocity of the block when it moves \(s=30 \mathrm{ft}\). The coefficient of kinetic friction at the surface is \(\mu_{s}=0.2\)

The ball of mass \(m\) is guided along the vertical circular path \(r=2 r_{c} \cos \theta\) using the arm \(O A\). If the arm has a constant angular velocity \(\dot{\theta}_{0}\), determine the angle \(\theta \leq 45^{\circ}\) at which the ball starts to leave the surface of the semicylinder. Neglect friction and the size of the ball. Probs. \(13-100 / 101\)

The \(300-\mathrm{kg}\) bar \(B\), originally at rest, is being towed over a series of small rollers. Determine the force in the cable when \(t=5 \mathrm{~s}\), if the motor \(M\) is drawing in the cable for a short time at a rate of \(v=\left(0.4 t^{2}\right) \mathrm{m} / \mathrm{s}\), where \(t\) is in seconds \((0 \leq t \leq 6 \mathrm{~s})\). How far does the bar move in \(5 \mathrm{~s}\) ? Neglect the mass of the cable, pulley, and the rollers. Prob. 13-38
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