/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 32 The uniform 225 -lb crate is sup... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 32

The uniform 225 -lb crate is supported by the thin homogeneous 40 -lb platform \(B F\) and light support links whose motion is controlled by the hydraulic cylinder \(C D .\) If the cylinder is extending at a constant rate of 6 in./sec when \(\theta=75^{\circ},\) determine the magnitudes of the forces supported by the pins at \(B\) and \(F\). Additionally, determine the total friction force acting on the crate. The crate is centered on the 6 -ft platform, and friction is sufficient to keep the crate motionless relative to the platform. (Hint: Be careful with the location of the resultant normal force beneath the crate.)

Short Answer

Expert verified
Reactions at pins B and F equal the total weight supported (265 lb combined), and total friction on the crate is zero.

Step by step solution

01

Analyze the System

Begin by considering the forces acting on the system. The system includes a 225-lb crate on a 40-lb platform with pins at points B and F, and a hydraulic cylinder at point D. The crate is in equilibrium on the platform, which means that all forces and moments must balance. The crate's location and platform length imply that the weight acts at the center of the platform.
02

Define Forces and Construct Free-Body Diagram

Define the forces at points B and F (reactions), the force exerted by the hydraulic cylinder at point D, and the weight located at the center of the platform. Construct a free-body diagram to visually represent these forces.
03

Calculate the Resultant Normal Force Location

Since the platform is homogeneous, the weight of the crate creates a uniform downward force. The normal force must be positioned to balance this. As the crate is centered, the normal force can be assumed to act at the midpoint unconstrained by any moment generated by platform or crate weight.
04

Write Equilibrium Equations

Use the conditions for equilibrium to establish sum of forces in the vertical direction and sum of moments around any point (typically either B or F for convenience):- For vertical forces: \( R_B + R_F = W_{crate} + W_{platform} \)- For moments: Take moments about B or F to eliminate one variable and solve for the other reaction first.
05

Solve for Reaction Forces at B and F

Substitute known values into your equilibrium equations. - Total weight: \( W = 225\text{ lb (crate)} + 40\text{ lb (platform)} = 265 \text{ lb} \)- If summing moments about B, solve for reaction at F, then substitute into the force equations to find the reaction at B.
06

Determine the Total Friction Force on the Crate

With the reactions known, the frictional force must balance any additional horizontal force exerted by the hydraulic cylinder. Since the crate is motionless, the frictional force is zero as there is no tendency for the crate to move horizontally.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Equilibrium
In dynamics, equilibrium refers to a state where all the forces and moments acting on an object are balanced. For an object to be in equilibrium, the sum of all forces in every direction must equate to zero. Moreover, the sum of all moments around any point must also equal zero.
This is crucial in ensuring that objects remain stationary or in constant motion, without acceleration. In the given exercise, the crate on the platform is in equilibrium, meaning the forces due to its weight and other interacting components like support pins and the hydraulic cylinder must perfectly counterbalance.
Analyzing equilibrium involves writing out the conditions for the sum of forces and the sum of moments. For example:
  • The vertical forces: The addition of the forces exerted by the support pins at B and F must equal the total weight which includes the crate and the platform.
  • The moments: Choosing a point (like B or F) to sum moments allows solving for unknowns systematically by eliminating variables.
Free-Body Diagram
A free-body diagram (FBD) is a graphical illustration that shows all external forces acting upon a system or body. It's a critical step in solving dynamics problems because it provides a clear view of how forces are distributed.
In our scenario, the FBD of the crate and platform system includes:
  • The weight of the crate and platform acting downward at the center.
  • The reaction forces at pins B and F acting upwards.
  • The force exerted by the hydraulic cylinder at point D.
Constructing a free-body diagram involves isolating each component and marking the force vectors. It's vital to keep track of all forces for accurate calculations. Use arrows to represent the direction of forces and label each force appropriately to follow through the problem-solving process.
Hydraulics
Hydraulics involves the use of fluid-based systems to create mechanical movement or force. In this exercise, the hydraulic cylinder controls the motion of the platform by extending at a constant speed.
The force exerted by the hydraulic cylinder must align with the equilibrium conditions for the system to remain steady. If the cylinder exerts too much or too little force, it could disrupt the equilibrium, causing movement or instability.
Understanding the role of hydraulics helps in predicting how changes in hydraulic pressure or cylinder extension rate can influence the system balance. From a design perspective, maintaining the correct hydraulic force ensures that the platform remains stable under the load of the crate.
Friction Force
Friction force is the resistance that one surface or object encounters when moving over another. It is crucial in stabilizing objects by preventing undesired motion.
In the exercise, friction ensures the crate remains stationary on the platform. Although the hydraulic cylinder moves, no horizontal force causes the crate to slide due to sufficient friction.
Friction relies on the contact between surfaces and can be static (preventing motion) or kinetic (opposing sliding motion). Here, the static friction is at play, balancing out any potential lateral forces, which can be explained by the equation:
  • The total frictional force equals zero as the crate is motionless and there is no horizontal motion.
Identifying frictional forces and their implications is crucial for understanding how systems maintain stability under varying forces.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Under active development is the storage of energy in high-speed rotating disks where friction is effectively eliminated by encasing the rotor in an evacuated enclosure and by using magnetic bearings. For a 10 -kg rotor with a radius of gyration of \(90 \mathrm{mm}\) rotating initially at \(80000 \mathrm{rev} / \mathrm{min},\) calculate the power \(P\) which can be extracted from the rotor by applying a constant \(2.10-\mathrm{N} \cdot \mathrm{m}\) retarding torque \((a)\) when the torque is first applied and ( \(b\) ) at the instant when the torque has been applied for 120 seconds.

The thin hoop of negligible mass and radius \(r\) contains a homogeneous semicylinder of mass \(m\) which is rigidly attached to the hoop and positioned such that its diametral face is vertical. The assembly is centered on the top of a cart of mass \(M\) which rolls freely on the horizontal surface. If the system is released from rest, what \(x\) -directed force \(P\) must be applied to the cart to keep the hoop and semicylinder stationary with respect to the cart, and what is the resulting acceleration \(a\) of the cart? Motion takes place in the \(x\) -y plane. Neglect the mass of the cart wheels and any friction in the wheel bearings. What is the requirement on the coefficient of static friction between the hoop and cart?

The uniform cylinder is rolling without slip with a velocity \(v\) along the horizontal surface when it overtakes a ramp traveling with speed \(v_{0} .\) Determine an expression for the speed \(v^{\prime}\) which the cylinder has relative to the ramp immediately after it rolls up onto the ramp. Finally, determine the percentage \(n\) of cylinder kinetic energy lost if \((a) \theta=10^{\circ}\) and \(v_{0}=0.25 v\) and (b) \(\theta=10^{\circ}\) and \(v_{0}=\) \(0.5 v .\) Assume that the clearance between the ramp and the ground is essentially zero, that the mass of the ramp is very large, and that the cylinder does not slip on the ramp.

The frame is made from uniform rod which has a mass \(\rho\) per unit length. A smooth recessed slot constrains the small rollers at \(A\) and \(B\) to travel horizontally. Force \(P\) is applied to the frame through a cable attached to an adjustable collar \(C .\) Determine the magnitudes and directions of the normal forces which act on the rollers if \((a) h=0.3 L,(b) h=0.5 L,\) and \((c) h=0.9 L .\) Evaluate your results for \(\rho=2 \mathrm{kg} / \mathrm{m}, L=500 \mathrm{mm},\) and \(P=60 \mathrm{N}\) What is the acceleration of the frame in each case?

The 1800 -kg rear-wheel-drive car accelerates forward at a rate of \(g / 2 .\) If the modulus of each of the rear and front springs is \(35 \mathrm{kN} / \mathrm{m},\) estimate the resulting momentary nose-up pitch angle \(\theta\). (This upward pitch angle during acceleration is called squat, while the downward pitch angle during braking is called dive!) Neglect the unsprung mass of the wheels and tires. (Hint: Begin by assuming a rigid vehicle.)
See all solutions

Recommended explanations on Physics Textbooks

Conservation of Energy and Momentum

Read Explanation

Nuclear Physics

Read Explanation

Classical Mechanics

Read Explanation

Waves Physics

Read Explanation

Rotational Dynamics

Read Explanation

Space Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.