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What minimum horizontal force \(F\) is needed to pull a wheel of radius \(R\) and mass \(M\) over a step of height \(h\) as shown in Fig. \(12-75(R>h) ?(a)\) Assume the force is applied at the top edge as shown. (b) Assume the force is applied instead at the wheel's center.

Short Answer

Expert verified
(a) The minimum force is \( F = \frac{Mg \cdot \cos\theta}{2} \). (b) The force is \( F = \frac{Mg(h-R)}{h} \).

Step by step solution

01

Analyze the Forces When Applied at Top Edge

When the force is applied at the top edge, the wheel is at the verge of punctual rotation over the step. We need to consider the torque about the point of contact at the step. The force of gravity acts downward at the center of the wheel. Torque due to the force \( F \) and torque due to gravitational force \( Mg \) must be equal for rotation to begin.
02

Calculate Torque Created by Force

Torque due to the applied force \( F \) at the top edge of the wheel is : \( \tau_F = F \cdot 2R \). This is because the perpendicular distance from the point of rotation is equal to \( 2R \).
03

Calculate Torque Created by Gravity

Torque created by gravitational force is given by \( \tau_{gravity} = Mg \cdot (R \cos\theta) \), where \( \theta \) is the angle formed when the wheel is about to rotate over the step.
04

Set Torques Equal and Solve for F

For the wheel to start rotating over the step, set \( \tau_F = \tau_{gravity} \). Therefore, \[ F \cdot 2R = Mg \cdot (R \cos\theta) \]. Solving for \( F \), we get \[ F = \frac{Mg \cdot \cos\theta}{2}. \]
05

Analyze Force Applied at the Center

When the force is applied at the center, the important aspect is the horizontal force that causes rotation. Calculate the horizontal distance which is leveraged by the applied force \( F \).
06

Torque Created by Central Force

Torque due to gravity acting at the center with a horizontally applied force is \( \tau_{gravity} = Mg \cdot (h-R) \), and the perpendicular distance is now \( h \).
07

Solve Torque Equation for Central Force

Use the equation \( F \cdot h = Mg(h-R) \). Solving for \( F \), we obtain \[ F = \frac{Mg(h-R)}{h} \].

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Key Concepts

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

Torque Calculation
Torque is a fundamental concept in physics when dealing with rotational forces. Simply put, torque is the measure of the rotational force applied at a distance from a pivot or fulcrum point. It's like a twisting force.
In our exercise, the wheel's rotation is considered around the contact point on the step, creating a system where torque due to the applied force and gravitational force must be understood.
The formula for torque is given by \( \tau = F \cdot d \), where \( F \) is the force applied and \( d \) is the perpendicular distance from the pivot point to the line of action of the force:
  • When the force is applied at the top of the wheel, the torque is \( \tau_F = F \cdot 2R \), since the force's line of action is twice the radius from the pivot.
  • This creates a balance where applied torque must match the counteracting torque from gravity for rotational motion to commence.
Understanding torque calculations helps in predicting scenarios like the wheel climbing over the step under different force applications.
Gravitational Force
Gravitational force is the attractive force exerted by the Earth on objects due to their mass. It plays a crucial role in rotational motion problems as it affects the object's weight, resulting in its tendency to rotate around a point.
When analyzing our wheel, gravitational force acts downward through the wheel’s center of mass. This force must be counteracted or harnessed correctly to achieve the desired motion.
In terms of torque caused by gravity, it’s computed with the component of gravitational force that acts tangentially. The expression \( \tau_{gravity} = Mg \cdot (R \cos\theta) \) involves:
  • \( Mg \) symbolizing the force due to the wheel's mass (M) and gravitational acceleration (g).
  • \( R \cos\theta \) as the effective perpendicular distance from the pivot when the wheel is on the verge of climbing over the step.
By setting gravitational torque equal to the applied torque, we can solve for the minimum force needed, appreciating the role gravity plays in manipulating rotational effects.
Rotational Motion
Rotational motion refers to the movement created when an object, like our wheel, spins around a pivot. Here, the wheel must rotate over a step which relies on balancing applied forces and gravitational torques.
The transition from static to rotational motion involves applying just enough force to start moving. Torque equations become pivotal, ensuring the wheel has enough moment to initiate rotational movement:
  • The wheel becomes a classic example of how rotation is influenced by external forces and gravitational effects.
  • Once the force is equal to or greater than the resisting gravitational torque at the pivot point, the wheel transitions to rotational motion over the step.
This concept reinforces the importance of understanding torque and force balance in practical physics applications, such as how objects begin rotating upon encountering obstacles.
Force Analysis
Force Analysis involves the methodical breakdown of all the forces acting on an object. It is essential for solving physics problems like determining how much force is needed to move the wheel over the obstacle step.
In the exercise, understanding how forces apply at different points impacts how the wheel overcomes the step:
  • When the force is applied at the top, the moment arm is longer, hence requiring a smaller force.
  • Applying force at the wheel's center changes dynamics, necessitating a higher force since the distance over which the force acts is reduced.
By evaluating different force applications, we can solve for the minimum force requirement in each scenario, unveiling insights into force influence on rotational systems. This step-by-step analysis simplifies the complexity inherent in rotational dynamics.

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Most popular questions from this chapter

(II) Three children are trying to balance on a seesaw, which includes a fulcrum rock acting as a pivot at the center, and a very light board \(3.2 \mathrm{~m}\) long (Fig. \(12-57\) ). Two playmates are already on either end. Boy \(A\) has a mass of \(45 \mathrm{~kg}\), and boy \(\mathrm{B}\) a mass of \(35 \mathrm{~kg} .\) Where should girl \(\mathrm{C},\) whose mass is \(25 \mathrm{~kg}\), place herself so as to balance the seesaw?

(II) A \(110-\mathrm{kg}\) horizontal beam is supported at each end. A \(320-\mathrm{kg}\) piano rests a quarter of the way from one end. What is the vertical force on each of the supports?

A uniform \(6.0-\mathrm{m}\) -long ladder of mass 16.0 \(\mathrm{kg}\) leans against a smooth wall (so the force exerted by the wall, \(\overline{\mathbf{F}}_{\mathrm{W}},\) is perpendicular to the wall). The ladder makes an angle of \(20.0^{\circ}\) with the vertical wall, and the ground is rough. Determine the coefficient of static friction at the base of the ladder if the ladder is not to slip when a 76.0 -kg person stands three-fourths of the way up the ladder.

(III) A cubic crate of side \(s=2.0 \mathrm{m}\) is top-heavy: its \(\infty \mathrm{is} 18 \mathrm{cm}\) above its true center. How steep an incline can the crate rest on without tipping over? What would your answer be if the crate were to slide at constant speed down the plane without tipping over? [Hint. The normal force would act at the lowest corner.]

A cube of side \(\ell\) rests on a rough floor. It is subjected to a steady horizontal pull \(F\), exerted a distance \(h\) above the floor as shown in Fig. \(12-84 .\) As \(F\) is increased, the block will either begin to slide, or begin to tip over. Determine the coefficient of static friction \(\mu_{\mathrm{s}}\) so that \((a)\) the block begins to slide rather than tip; \((b)\) the block begins to tip. [Hint: Where will the normal force on the block act if it tips?

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