/*! 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} Q16P Question: A uniform cubical crat... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Question: A uniform cubical crate is 0.750 m on each side and weighs 500 N . It rests on a floor with one edge against a very small, fixed obstruction. At what least height above the floor must a horizontal force of magnitude 350 N be applied to the crate to tip it?

Short Answer

Expert verified

Answer:

A horizontal force of 350 N must be applied at to tip the crate.

Step by step solution

01

Understanding the given information   

a=0.75mW=500NF=350N

02

Concept and formula used in the given question

Using the conditions for static equilibrium, you can write the equation for torque in terms of force and distance. Using this equation, you can solve for unknown height h. The equations used are given below.

Static equilibrium conditions:

∑Fx=0∑Fy=0∑τ=0

03

Calculation for the at least what height above the floor must a horizontal force of magnitude  350 N be applied to the crate to tip it  

Applying equilibrium condition:

∑τ=0W×a2-F×h=0500×0.752-350×h=0h=0.536m

Hence,a horizontal force of 350 N must be applied ath=0.536m to tip the crate.

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 91Ó°ÊÓ!

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

The system in Fig. 12-77 is in equilibrium. The angles are θ1=60°and θ2=20°, and the ball has mass M=2.0 k²µ. What is the tension in (a) string ab and (b) string bc?

In Fig. 12-63, a rectangular slab of slate rests on a bedrock surface inclined at angle θ=26°. The slab has length L=43 m, thickness T=2.5 m, and width,W=12 mand 1.0 cm3of it has a mass of 3.2g. The coefficient of static friction between slab and bedrock is 0.39. (a) Calculate the component of the gravitational force on the slab parallel to the bedrock surface. (b) Calculate the magnitude of the static frictional force on the slab. By comparing (a) and (b), you can see that the slab is in danger of sliding. This is prevented only by chance protrusions of bedrock. (c) To stabilize the slab, bolts are to be driven perpendicular to the bedrock surface (two bolts are shown). If each bolt has a cross-sectional area of 6.4 c³¾2and will snap under a shearing stress of, 3.6×108N/m2what is the minimum number of bolts needed? Assume that the bolts do not affect the normal force.

Figure 12-85ashows details of a finger in the crimp holdof the climber in Fig. 12-50. A tendon that runs from muscles inthe forearm is attached to the far bone in the finger. Along the way, the tendon runs through several guiding sheaths called pulleys. The A2 pulley is attached to the first finger bone; the A4 pulley is attached to the second finger bone. To pull the finger toward the palm, the forearm muscles pull the tendon through the pulleys, much like strings on a marionette can be pulled to move parts of the marionette. Figure 12-85bis a simplified diagram of the second finger bone, which has length d. The tendon’s pull F→ton the bone acts at the point where the tendon enters the A4 pulley, at distance d/3 along the bone. If the force components on each of the four crimped fingers in Fig. 12-50 are Fh=13.4 Nand Fv=162.4 N, what is the magnitude ofF→t ? The result is probably tolerable, but if the climber hangs by only one or two fingers, the A2 and A4 pulleys can be ruptured, a common ailment among rock climbers.

In Fig.12-48, the driver of a car on a horizontal road makes an emergency stop by applying the brakes so that all four wheels lock and skid along the road. The coefficient of kinetic friction between tires and road is 0.40.The separation between the front and rear axles is L=4.2 m, and the center of mass of the car is located at distance d=1.8 m behind the front axle and distanceh=0.75 m above the road. Theweight of car is 11 kN. Find the magnitude of (a) the braking acceleration of the car, (b) the normal force on each rear wheel, (c) the normal force on each front wheel, (d) the braking force on each rear wheel, and (e) the braking force on each front wheel. (Hint:Although the car is not in translational equilibrium, it isin rotational equilibrium.)

A pan balance is made up of a rigid, massless rod with a hanging pan attached at each end. The rod is supported at and free to rotate about a point not at its center. It is balanced by unequal masses placed in the two pans. When an unknown mass mis placed in the left pan, it is balanced by a mass m1 placed in the right pan; when the mass mis placed in the right pan, it is balanced by a mass m2in the left pan. Show thatm=m1m2

See all solutions

Recommended explanations on Physics Textbooks

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.