College Physics
OER 2016 Edition
ISBN: 9781947172173
Author: OpenStax
Publisher: OpenStax College
expand_more
expand_more
format_list_bulleted
Concept explainers
Videos
Question
Chapter 9, Problem 14PE
To determine
(a)
The tension in the chain.
To determine
(b)
The force exerted by each side on the hinge.
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
Hi, I am in physics I and I'm trying to understand this word problem. It saids that a 6.0-m uniform board is supported by two sawhorses 4.0m apart. A 32kg child walks on the board to 1.4 m beyond the right support when the board start to tip.
a.) Find the mass of the board
b.) Then he walks back 0.40m and stands still. Find the forces exerted by the supports at this time. (use the mass found in part a).
Please be detailed of how this works. Its a harder problem for me to understand. Thank you!
A homeowner is trying to move a stubborn rock from his yard. By using a a metal rod as a lever arm and a fulcrum (or pivot point) the homeowner will have a better chance of moving the rock. The homeowner places the fulcrum a distance ?=0.288 mfrom the rock, which has a mass of 465 kg, and fits one end of the rod under the rock's center of weight.
If the homeowner can apply a maximum force of 671 N at the other end of the rod, what is the minimum total length ? of the rod required to move the rock? Assume that the rod is massless and nearly horizontal so that the weight of the rock and homeowner's force are both essentially vertical. The acceleration due to gravity is ?=9.81 m/s2.
An individual leans forwards to pick up a box of 100 N. The weight of his upper body has a magnitude of 450 N. The back is pivoting around the base of the vertebral column. Consider the back of the individual as a rigid bar that is controlled by a muscle with an angle of 12° (See picture, d = trunk-head distance = 1 m).a) Calculate the magnitude of muscle force required to lift the box.b) Calculate the magnitude of the force at the base of the vertebral column. Hints: For (a) solve the equilibrium of moments, i.e. what force is required in the muscle to balance out the moments acting around the base of the spine.For (b), solve the equilibrium of forces acting on the spine, including the muscle force you’ve just calculated, in x and y separately. There are two extra forces not shown in the diagram: x and y contact forces acting at the base of the spine. These are whatever is needed to keep the total forces acting on the spine = 0 (so the spine isn’t accelerating off in some…
Chapter 9 Solutions
College Physics
Ch. 9 - What can you say about the velocity of a moving...Ch. 9 - Under what conditions can a rotating body be in...Ch. 9 - What three factors affect the torque created by a...Ch. 9 - A wrecking ball is being used to knock down a...Ch. 9 - Mechanics put a length of Pipe over the handle of...Ch. 9 - Prob. 6CQCh. 9 - Explain the need for tall towers on a suspension...Ch. 9 - When visiting some countries, you may see a person...Ch. 9 - Prob. 9CQCh. 9 - Prob. 10CQ
Ch. 9 - Why are the forces exerted on the outside world by...Ch. 9 - Explain why the forces in our joints are several...Ch. 9 - Why are the forces exerted on the outside world by...Ch. 9 - Explain why the forces in our joints are several...Ch. 9 - Certain of dinosaurs were bipedal (walked on two...Ch. 9 - Swimmers and athletes during competition need to...Ch. 9 - If the maximum force the biceps muscle can exert...Ch. 9 - Suppose the biceps muscle was attached through...Ch. 9 - Explain one of the reasons why pregnant women...Ch. 9 - (a) When opening a door, you push on it...Ch. 9 - When tightening a bolt, you push perpendicularly...Ch. 9 - Two children push on opposite sides of a door...Ch. 9 - Use the second condition for equilibrium (net =0 )...Ch. 9 - Repeat the seesaw problem in Example 9.1 with the...Ch. 9 - Prob. 6PECh. 9 - Two children of mass 20.0 kg and 30.0 kg sit...Ch. 9 - Prob. 8PECh. 9 - A person carries a plank of wood 2.00 m long with...Ch. 9 - Prob. 10PECh. 9 - Prob. 11PECh. 9 - Prob. 12PECh. 9 - Prob. 13PECh. 9 - Prob. 14PECh. 9 - Prob. 15PECh. 9 - Prob. 16PECh. 9 - To get up on the roof, a person (mass 70.0 kg)...Ch. 9 - Prob. 18PECh. 9 - Prob. 19PECh. 9 - Suppose you needed to raise a 250-kg mower a...Ch. 9 - Prob. 21PECh. 9 - Prob. 22PECh. 9 - Prob. 23PECh. 9 - Prob. 24PECh. 9 - Prob. 25PECh. 9 - Prob. 26PECh. 9 - Prob. 27PECh. 9 - Prob. 28PECh. 9 - Prob. 29PECh. 9 - Prob. 30PECh. 9 - Prob. 31PECh. 9 - Prob. 32PECh. 9 - Prob. 33PECh. 9 - Prob. 34PECh. 9 - Prob. 35PECh. 9 - Integrated Concepts Suppose we replace the 4.0-kg...Ch. 9 - Prob. 37PECh. 9 - You have just planted a sturdy 2-m-tall palm tree...Ch. 9 - Unreasonable Results Suppose two children are...Ch. 9 - Construct Your Own Problem Consider a method for...Ch. 9 - Prob. 1TPCh. 9 - Prob. 2TPCh. 9 - Prob. 3TPCh. 9 - Prob. 4TPCh. 9 - Prob. 5TPCh. 9 - Prob. 6TPCh. 9 - Prob. 7TPCh. 9 - Prob. 8TPCh. 9 - Prob. 9TPCh. 9 - Prob. 10TPCh. 9 - Prob. 11TP
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.Similar questions
- A stepladder of negligible weight is constructed as shown in Figure P10.73, with AC = BC = = 4.00 m. A painter of mass m = 70.0 kg stands on the ladder d = 3.00 m from the bottom. Assuming the floor is frictionless, find (a) the tension in the horizontal bar DE connecting the two halves of the ladder, (b) the normal forces at A and B, and (c) the components of the reaction force at the single hinge C that the left half of the ladder exerts on the right half. Suggestion: Treat the ladder as a single object, but also treat each half of the ladder separately.arrow_forwardA stepladder of negligible weight is constructed as shown in Figure P10.73, with AC = BC = ℓ. A painter of mass m stands on the ladder a distance d from the bottom. Assuming the floor is frictionless, find (a) the tension in the horizontal bar DE connecting the two halves of the ladder, (b) the normal forces at A and B, and (c) the components of the reaction force at the single hinge C that the left half of the ladder exerts on the right half. Suggestion: Treat the ladder as a single object, but also treat each half of the ladder separately. Figure P10.73 Problems 73 and 74.arrow_forwardA wooden door 2.1 m high and 0.90 m wide is hung by two hinges 1.8 m apart. The lower hinge is 15 cm above the bottom of the door. The center of mass of the door is at its geometric center, and the weight of the door is 260 N, which is supported equally by both hinges. Find the horizontal force exerted by each hinge on the door.arrow_forward
- A stepladder of negligible weight is constructed as shown in Figure P12.40, with AC = BC = = 4.00 m. A painter of mass m = 70.0 kg stands on the ladder d = 3.00 m from the bottom. Assuming the floor is frictionless, find (a) the tension in the horizontal bar DE connecting the two halves of the ladder, (b) the normal forces at A and B, and (c) the components of the reaction force at the single hinge C that the left half of the ladder exerts on the right half. Suggestion: Treat the ladder as a single object, but also treat each half of the ladder separately. Figure P12.40 Problems 40 and 41.arrow_forwardRuby, with mass 55.0 kg, is trying to reach a box on a high shelf by standing on her tiptoes. In this position, half her weight is supported by the normal force exerted by the floor on the toes of each foot as shown in Figure P14.75A. This situation can be modeled mechanically by representing the force on Rubys Achilles tendon with FA and the force on her tibia as FT as shown in Figure P14.75B. What is the value of the angle and the magnitudes of the forces FA and FT? FIGURE P14.75arrow_forward(a) What minimum coefficient of friction is needed between the legs and the ground to keep the sign in Figure 9.35 in the position shown if the chain breaks? (b) What force is exerted by each side on the hinge?arrow_forward
- A uniform beam resting on two pivots has a length L = 6.00 m and mass M = 90.0 kg. The pivot under the left end exerts a normal force n1 on the beam, and the second pivot located a distance = 4.00 m from the left end exerts a normal force n2. A woman of mass m = 55.0 kg steps onto the left end of the beam and begins walking to the right as in Figure P10.28. The goal is to find the womans position when the beam begins to tip. (a) What is the appropriate analysis model for the beam before it begins to tip? (b) Sketch a force diagram for the beam, labeling the gravitational and normal forces acting on the beam and placing the woman a distance x to the right of the first pivot, which is the origin. (c) Where is the woman when the normal force n1 is the greatest? (d) What is n1 when the beam is about to tip? (e) Use Equation 10.27 to find the value of n2 when the beam is about to tip. (f) Using the result of part (d) and Equation 10.28, with torques computed around the second pivot, find the womans position x when the beam is about to tip. (g) Check the answer to part (e) by computing torques around the first pivot point. Figure P10.28arrow_forwardAt a museum, a 1300-kg model aircraft is hung from a lightweight beam of length 12.0 m that is free to pivot about its base and is supported by a massless cable (Fig. P14.38). Ignore the mass of the beam. a. What is the tension in the section of the cable between the beam and the wall? b. What are the horizontal and vertical forces that the pivot exerts on the beam? FIGURE P14.38 (a) From the free-body diagram, the angle that the string tension makes with the beam is = 55.0 + 18.0 = 73.0, and the perpendicular component of the string tension is FT sin73.0. Summing torques around the base of the rod gives (Eq. 14.2): =0:(12.0m)(1300kg)(9.81m/s2)cos55.0+FT(12.0m)sin73.0=0FT=(12.0m)(1300kg)(9.81m/s2)cos55.0(12.0m)sin73.0FT=7.65103N Figure P14.38ANS (b) Using force balance (Eq. 14.1): Fx=0:FHFTcos18.0=0FH=FTcos18.0=[(12.0m)(1300kg)(9.81m/s2)cos55.0(12.0m)sin73.0]cos18.0=7.27103NFy=0:FVFTsin18.0(1300kg)(9.81m/s2)=0 FV=FTsin18.0+(1300kg)gFV=[(12.0m)(1300kg)(9.81m/s2)cos55.0(12.0m)sin73.0]sin18.0+(1300kg)(9.81m/s2)FV=1.51104Narrow_forwardIn order to get his car out of the mud, a man ties one end of a rope to the front bumper and the other end to a tee 15 m away as shown below. He then pulls on the center of the rope with a force of 400 N, which causes its center to be displaced 0.30 m, as shown. What is the force of the rope on the car?arrow_forward
- A massless, horizontal beam of length L and a massless rope support a sign of mass m (Fig. P14.78). a. What is the tension in the rope? b. In terms of m, g, d, L, and , what are the components of the force exerted by the beam on the wall? FIGURE P14.78arrow_forwarda) What is the mechanical advantage of a wheelbarrow, such as the one in Figure 9.24, if the center of gravity of the wheelbarrow and its load has a perpendicular lever arm of 5.50 cm, while the hands have a perpendicular lever arm of 1.02 m? (b) What upward force should you exert to support the wheelbarrow and its load if their combined mass is 55.0 kg? (c) What force does the wheel exert on the ground? Figure 9.24 (a) In the case of the wheelbarrow, the output force or load is between the pivot and the input force. The pivot IS the wheel's axle. Here, the output force is greater than the input force. Thus, a wheelbarrow enables you to lift much heavier loads than you could with your body alone. (b) In the case of the shovel, the input force is between the pivot and the load, but the input lever arm is shorter than the output lever arm. The pivot is at the handle held by the right hand. Here, the output force (supporting the shovel's load) is less than the input force (from the hand nearest the load), because the Input is exerted closer to the pivot than is the outputarrow_forwardUnder what conditions can a rotating body be in equilibrium? Give an example.arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
Principles of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning
Physics for Scientists and EngineersPhysicsISBN:9781337553278Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning
Physics for Scientists and Engineers with Modern ...PhysicsISBN:9781337553292Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning
Physics for Scientists and Engineers: Foundations...PhysicsISBN:9781133939146Author:Katz, Debora M.Publisher:Cengage Learning
College PhysicsPhysicsISBN:9781938168000Author:Paul Peter Urone, Roger HinrichsPublisher:OpenStax College
Physics for Scientists and Engineers, Technology ...PhysicsISBN:9781305116399Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning
Principles of Physics: A Calculus-Based Text
Physics
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Physics for Scientists and Engineers
Physics
ISBN:9781337553278
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Physics for Scientists and Engineers with Modern ...
Physics
ISBN:9781337553292
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Physics for Scientists and Engineers: Foundations...
Physics
ISBN:9781133939146
Author:Katz, Debora M.
Publisher:Cengage Learning
College Physics
Physics
ISBN:9781938168000
Author:Paul Peter Urone, Roger Hinrichs
Publisher:OpenStax College
Physics for Scientists and Engineers, Technology ...
Physics
ISBN:9781305116399
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Static Equilibrium: concept; Author: Jennifer Cash;https://www.youtube.com/watch?v=0BIgFKVnlBU;License: Standard YouTube License, CC-BY