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Chapter 4: Problem 55

(III) A small block of mass \(m\) rests on the sloping side of a triangular block of mass \(M\) which itself rests on a hori- zontal table as shown in Fig. \(47 .\) Assuming all surfaces are frictionless, determine the magnitude of the force \(\vec{\mathbf{F}}\) that must be applied to \(M\) so that \(m\) remains in a fixed position relative to \(M\) (that is, \(m\) doesn't move on the incline). [Hint: Take \(x\) and \(y\) axes horizontal and vertical.

Short Answer

Expert verified
The force \( \vec{F} = (M + m)g \tan(\theta) \) keeps block \( m \) stationary on the incline.

Step by step solution

01

Analyze the system

The system consists of two blocks, a smaller block of mass \( m \) on an incline of a larger block of mass \( M \), which itself is on a frictionless horizontal table. We need a force \( \vec{F} \) on the larger block to keep the smaller block stationary on the inclined plane. Since all surfaces are frictionless, the force \( \vec{F} \) causes the entire system to undergo horizontal motion.
02

Set up forces on the small block

For the small block \( m \) to remain stationary relative to block \( M \), the horizontal and vertical components of any force that acts on it must balance. The forces acting on \( m \) are its weight \( \mathbf{W} = m \mathbf{g} \) acting down, and a normal force \( \mathbf{N} \) from the inclined surface.
03

Resolve weight components

Resolve the weight \( \mathbf{W} = m \mathbf{g} \) into components parallel and perpendicular to the incline. If \( \theta \) is the angle of the incline, then:- Parallel to the incline: \( W_{ ext{parallel}} = m g \sin(\theta) \)- Perpendicular to the incline: \( W_{ ext{perpendicular}} = m g \cos(\theta) \)
04

Establish conditions for no movement

For the block \( m \) to not move on the incline, the net force along the incline should be zero. Therefore, any horizontal acceleration \( a \) of the system must be such that the horizontal component of the normal force offsets the component of gravitational force parallel to the incline.
05

Calculate the necessary acceleration

For the block \( m \) to have no relative motion on \( M \), the acceleration \( a \) of the entire system must equal the acceleration caused by the gravitational component along the plane:\[ a = g \tan(\theta) \]
06

Determine the force \( \vec{F} \)

With \( a = g \tan(\theta) \), the force \( \vec{F} \) needed to produce this acceleration on block \( M \) of mass \( M \) (plus the mass \( m \) on it) is calculated as:\[ \vec{F} = (M + m)g \tan(\theta) \]

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Key Concepts

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

Newton's Laws of Motion
Newton's Laws of Motion provide a framework to understand how forces cause motion. There are three essential laws:
  • First Law (Law of Inertia): An object remains at rest or in uniform motion unless acted upon by a net external force.
  • Second Law: The acceleration of an object is proportional to the net force acting on it and inversely proportional to its mass, represented by the equation: \( \vec{F} = m \vec{a} \). This is crucial for our frictionless incline problem, as it tells us how the force \( \vec{F} \) applied to block \( M \) causes the entire system to accelerate.
  • Third Law: For every action, there is an equal and opposite reaction.
In our exercise, the second law is particularly important. It explains why the force \( \vec{F} \) applied on the larger block causes both blocks (\( M \) and \( m \)) to move together, ensuring that the smaller block doesn't slide down the incline. By analyzing forces and resulting motions using these laws, we can determine the precise force needed to keep the block \( m \) stationary relative to \( M \).
Inclined Plane Mechanics
An inclined plane is a flat surface tilted at an angle to the horizontal. Understanding inclined plane mechanics involves analyzing how forces interact when an object is placed on such a surface.

In our problem, the smaller block rests on an inclined surface that's part of the larger block \( M \). Forces acting on the small block include its weight \( m \mathbf{g} \) and the normal force from the incline. These forces must be resolved into components
  • Parallel to the incline: \( m g \sin(\theta) \)
  • Perpendicular to the incline: \( m g \cos(\theta) \)
These components help us understand how the forces keep the block from sliding down. The horizontal component is directly counteracted by the horizontal movement of \( M \), thanks to the force \( \vec{F} \) that causes the entire system to accelerate together.

Thus, inclined plane mechanics helps explain why balancing these components is key to keeping the smaller block from moving relative to the inclined surface.
Normal Force Calculation
The normal force is the perpendicular force exerted by a surface on an object resting on it. In our problem, it is essential to determine the normal force acting on the smaller block to prevent it from sliding down the incline.

The normal force \( \mathbf{N} \) acts perpendicular to the inclined surface and balances the perpendicular component of the block's weight.
  • The component of weight perpendicular to the incline is \( m g \cos(\theta) \).
  • Thus, the normal force \( \mathbf{N} \) is equal in magnitude to this component, ensuring no vertical motion.
However, since the entire system undergoes horizontal acceleration due to the force \( \vec{F} \), \( \mathbf{N} \) plays an additional crucial role. Its horizontal component must balance the parallel component of weight \( m g \sin(\theta) \) to maintain no relative motion between \( m \) and \( M \).

Therefore, understanding normal force calculation helps us ensure the correct conditions are met to keep the block stationary on the incline during the system's acceleration.

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Most popular questions from this chapter

(II) A fisherman yanks a fish vertically out of the water with an acceleration of 2.5 \(\mathrm{m} / \mathrm{s}^{2}\) using very light fishing line that has a breaking strength of 18 \(\mathrm{N}(\approx 4\) lb). The fisherman unfortunately loses the fish as the line snaps. What can you say about the mass of the fish?

(II) High-speed elevators function under two limitations: \((1)\) the maximum magnitude of vertical acceleration that a typical human body can experience without discomfort is about \(1.2 \mathrm{m} / \mathrm{s}^{2},\) and \((2)\) the typical maximum speed attainable is about 9.0 \(\mathrm{m} / \mathrm{s}\) . You board an elevator on a skyscraper's ground floor and are transported 180 \(\mathrm{m}\) above the ground level in three steps: acceleration of magnitude 1.2 \(\mathrm{m} / \mathrm{s}^{2}\) from rest to 9.0 \(\mathrm{m} / \mathrm{s}\) , followed by constant upward velocity of 9.0 \(\mathrm{m} / \mathrm{s}\) , then deceleration of magnitude 1.2 \(\mathrm{m} / \mathrm{s}^{2}\) from 9.0 \(\mathrm{m} / \mathrm{s}\) to rest. (a) Determine the elapsed time for each of these 3 stages. \((b)\) Determine the change in the magnitude of the normal force, expressed as a \(\%\) of your normal weight during each stage, (c) What fraction of the total transport time does the normal force not equal the person's weight?

A skateboarder, with an initial speed of \(2.0 \mathrm{~m} / \mathrm{s}\), rolls virtually friction free down a straight incline of length \(18 \mathrm{~m}\) in \(3.3 \mathrm{~s}\). At what angle \(\theta\) is the incline oriented above the horizontal?

(I) Sketch the free-body diagram of a baseball (a) at the moment it is hit by the bat, and again \((b)\) after it has left the bat and is flying toward the outfield.

(III) A heavy steel cable of length \(\ell\) and mass \(M\) passes over a small massless, frictionless pulley. (a) If a length \(y\) hangs on one side of the pulley (so \(\ell-y\) hangs on the other side), calculate the acceleration of the cable as a function of \(y .\) (b) Assuming the cable starts from rest with length \(y_{0}\) on one side of the pulley, determine the velocity \(v_{\mathrm{r}}\) at the moment the whole cable has fallen from the pulley. (c) Evaluate \(v_{\mathrm{f}}\) for \(y_{0}=\frac{2}{3} \ell\) . [Hint: Use the chain rule, \(d v / d t=(d v / d y)(d y / d t),\) and integrate. \(]\)
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