Sample Problem 12.02 Balancing a leaning boomFigure 12-6a shows a safe (mass M = 430 kg) hanging by arope (negligible mass) from a boom (a = 1.9 m and b =2.5 m) that consists of a uniform hinged beam (m = 85 kg)and horizontal cable (negligible mass).Figure 12-6 (a)A heavy safe ishung from aboom consistingCableof a horizontal(a) What is the tension T, in the cable? In other words, what isthe magnitude of the force T on the beam from the cable?steel cable and auniform beam.Beam comRope(b) A free-bodydiagram for thebeam.KEY IDEASThe system here is the beam alone, and the forces on it areshown in the free-body diagram of Fig. 12-6b. The forcefrom the cable is T. The gravitational force on the beamacts at the beam's center of mass (at the beam's center) andis represented by its equivalent mg. The vertical componentof the force on the beam from the hinge is F, and the hori-zontal component of the force from the hinge is F. Theforce from the rope supporting the safe is T. Because beam,rope, and safe are stationary, the magnitude of T, is equal tothe weight of the safe: T, = Mg. We place the origin O of anxy coordinate system at the hinge. Because the system is instatic equilibrium, the balancing equations apply to it.Hinge(a)MBeammgHere is theF,wise choice ofCalculations: Let us start with Eq. 12-9 (Tnet.z = 0). Notethat we are asked for the magnitude of force T, and not offorces F, and F, acting at the hinge, at point O. To eliminateF, and F, from the torque calculation, we should calculatetorques about an axis that is perpendicular to the figure atpoint O. Then F, and F will have moment arms of zero. Thelines of action for T, T, and mg are dashed in Fig. 12-6b.The corresponding moment arms are a, b, and b/2.Writing torques in the form of r Fand using our rule aboutsigns for torques, the balancing equation Tet.z = 0 becomesrotation axis.(b)Fmagnitude F of the net force. Because we know T, we applythe force balancing equations to the beam.Calculations: For the horizontal balance, we can rewriteFst = 0 asF, – T = 0,(12-20)(a)(T.) – (b)(T,) – (¿b)(mg) = 0.(12-19)and soF, = T = 6093 N.Substituting Mg for T, and solving for T, we find thatFor the vertical balance, we write Fety = 0 asgb(M + m)F, - mg – T, = 0.T =aSubstituting Mg for T, and solving for F, we find that(9.8 m/s²)(2.5 m)(430 kg + 85/2 kg)F, = (m + M)g = (85 kg + 430 kg)(9.8 m/s²)1.9 m= 6093 N = 6100 N.(Answer)= 5047 N.From the Pythagorean theorem, we now have(b) Find the magnitude F of the net force on the beam fromthe hinge.F = VF} + F2= V(6093 N)? + (5047 N)² = 7900 N.(Answer)KEY IDEANow we want the horizontal component F, and verticalcomponent F, so that we can combine them to get theNote that Fis substantially greater than either the combinedweights of the safe and the beam, 5000 N, or the tension inthe horizontal wire, 6100 N.PLUS Additional examples, video, and practice available at WileyPLUS In sample problem 12.02, the vertical forceacting on the beam by the hinge is equal toа. Mgb. Tr+ mgС.Trd. Tc

Answered: Image /qna-images/answer/7bba0b9a-b977-48c6-820d-d4162c4776ea.jpg

Answered: In sample problem 12.02, the vertical…

Science

Physics

In sample problem 12.02, the vertical force acting on the beam by the hinge is equal to а. Mg b. Tr+ mg С. Tr d. To

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

See similar textbooks

Related questions

Q: steel beam of uniform cross section weighs 2.50 x 10^5 N. if it is 5.00m long. what force is…

Q: A 7.64-m uniform beam is hinged to a vertical wall and held horizontally by a 5.00-m cable attached…

Q: One end of a uniform = 4.10-m-long rod of weight w is supported by a cable at an angle of 8 = 37°…

Q: A solid, uniform plank is held with two supports each providing a normal force at the ends. The…

Q: 10 m F A hinged beam is connected to a wall (see Figure). The opposite end rests on the ground. If a…

Q: 1. What force must be applied to the bar represented in the figure, for it to be in balance? L = 2m…

Q: A window washer is standing on a scaffold supported by a vertical rope at each end. The scaffold…

Q: EXT atol He oll The sign on the building is attached to the wall by a hinge and a cable. It has a…

A: Consider the blue portion as area 1 and green portion as area 2.

Q: force F of 30 N is applied to a door. Find the torque if F is applied to the door: a) 0.80 m from…

Q: A glass optical fiber in a communications system has a diameter of 9.0 μm. a. What maximum force…

A: Given:Diameter of the optical fiber(d) is 9.0μmUsing the formula:As the maximum stress of the object…

Q: Problems 1, 2, and 3 refer to the following figure and situation. A uniform beam of mass m = 19.4 kg…

Q: SERPSE10 12.3.P.006. A uniform beam of length L = 7.70 m and weight 4.90 x 10² N is carried by two…

A: According to newton's Second law Fsam+Fjoe−w=0 Fsam=490−Fjoe equation 1

Q: I need help with this question, please

Q: A young work crew is trying to quickly finish some touch-up painting to complete a job. Rather than…

A: For the plank to remain motionless, net torque on it should be zero. So we will first write an…

Q: 11. Arectangular sign of mass 50.0 kg and width w = 5.00 m and e = length 4.00 m is hanging from a…

A: Write the given values using suitable variables. Here m signifies rectangular mass, w signifies…

Q: Typewritten only and i'll upvote. The uniform horizontal beam supporting the sign is 1.39 m long,…

A: Solution:-Given that

Q: The sign on the building is attached to the wall by a hinge and a cable. It has a mass of 240 kg.…

A: Consider the blue portion as area 1 and green portion as area 2.

Q: A 14.0 m uniform ladder weighing 490 N rests against a frictionless wall. The ladder makes a…

Q: 13. A uniform 8.00 m, 580 N beam is hinged to a wall and supported by a thin cable attached 2.00 m…

Q: For the door shown below, find the forces in unit vector notation on the (a) The top hinge 0.28 m…

A: Dear student If any rigid body is in equilibrium, the net force as well as net torque on it is…

Q: AB = 5008.95 Ib AE = -4407.78 Ib BC = -2101.41 Ib BD = -7882.09 Ib BE = 2910 Ib CD = 4485.98 Ib DE =…

Q: An office wants to hang a sign, mass 90 kg, which is supported by a uniform 490 N beam as shown in…

Q: As shown below, a 258 N horizontal beam is in equilibrium. The orange circle represents the hinge.…

A: The forces acting are shown in the firgure below. In equilibrium, the total moment about the hinge…

Q: A slightly simplified view of the arm has the hinge at the elbow, the bicep attaching to the bone…

Q: uniform horizontal strut weighs 340.0 N. One end of the strut is attached to a hinged support at the…

A: Taking the sum of moments about the left end, in the hinged support:Here, T is the tension in the…

Q: Please don't provide handwritten solution...

A: Solution To solve the problem of balancing the seesaw, we need to use the concept of torque…

Concept explainers

Rotational Equilibrium And Rotational Dynamics

In physics, the state of balance between the forces and the dynamics of motion is called the equilibrium state. The balance between various forces acting on a system in a rotational motion is called rotational equilibrium or rotational dynamics.

Equilibrium of Forces

The tension created on one body during push or pull is known as force.

Question

Sample Problem 12.02 Balancing a leaning boom
Figure 12-6a shows a safe (mass M = 430 kg) hanging by a
rope (negligible mass) from a boom (a = 1.9 m and b =
2.5 m) that consists of a uniform hinged beam (m = 85 kg)
and horizontal cable (negligible mass).
Figure 12-6 (a)
A heavy safe is
hung from a
boom consisting
Cable
of a horizontal
(a) What is the tension T, in the cable? In other words, what is
the magnitude of the force T on the beam from the cable?
steel cable and a
uniform beam.
Beam com
Rope
(b) A free-body
diagram for the
beam.
KEY IDEAS
The system here is the beam alone, and the forces on it are
shown in the free-body diagram of Fig. 12-6b. The force
from the cable is T. The gravitational force on the beam
acts at the beam's center of mass (at the beam's center) and
is represented by its equivalent mg. The vertical component
of the force on the beam from the hinge is F, and the hori-
zontal component of the force from the hinge is F. The
force from the rope supporting the safe is T. Because beam,
rope, and safe are stationary, the magnitude of T, is equal to
the weight of the safe: T, = Mg. We place the origin O of an
xy coordinate system at the hinge. Because the system is in
static equilibrium, the balancing equations apply to it.
Hinge
(a)
M
Beam
mg
Here is the
F,
wise choice of
Calculations: Let us start with Eq. 12-9 (Tnet.z = 0). Note
that we are asked for the magnitude of force T, and not of
forces F, and F, acting at the hinge, at point O. To eliminate
F, and F, from the torque calculation, we should calculate
torques about an axis that is perpendicular to the figure at
point O. Then F, and F will have moment arms of zero. The
lines of action for T, T, and mg are dashed in Fig. 12-6b.
The corresponding moment arms are a, b, and b/2.
Writing torques in the form of r Fand using our rule about
signs for torques, the balancing equation Tet.z = 0 becomes
rotation axis.
(b)
F
magnitude F of the net force. Because we know T, we apply
the force balancing equations to the beam.
Calculations: For the horizontal balance, we can rewrite
Fst = 0 as
F, – T = 0,
(12-20)
(a)(T.) – (b)(T,) – (¿b)(mg) = 0.
(12-19)
and so
F, = T = 6093 N.
Substituting Mg for T, and solving for T, we find that
For the vertical balance, we write Fety = 0 as
gb(M + m)
F, - mg – T, = 0.
T =
a
Substituting Mg for T, and solving for F, we find that
(9.8 m/s²)(2.5 m)(430 kg + 85/2 kg)
F, = (m + M)g = (85 kg + 430 kg)(9.8 m/s²)
1.9 m
= 6093 N = 6100 N.
(Answer)
= 5047 N.
From the Pythagorean theorem, we now have
(b) Find the magnitude F of the net force on the beam from
the hinge.
F = VF} + F2
= V(6093 N)? + (5047 N)² = 7900 N.
(Answer)
KEY IDEA
Now we want the horizontal component F, and vertical
component F, so that we can combine them to get the
Note that Fis substantially greater than either the combined
weights of the safe and the beam, 5000 N, or the tension in
the horizontal wire, 6100 N.
PLUS Additional examples, video, and practice available at WileyPLUS

In sample problem 12.02, the vertical force
acting on the beam by the hinge is equal to
а. Mg
b. Tr+ mg
С.
Tr
d. Tc

Expert Solution

This question has been solved!

Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.

This is a popular solution!

SEE SOLUTION Check out a sample Q&A here

Step 1

VIEW

Step 2

VIEW

Trending now

This is a popular solution!

Step by step

Solved in 2 steps with 2 images

SEE SOLUTION Check out a sample Q&A here

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

Needs Complete typed solution with 100 % accuracy.
Needs Complete typed solution with 100 % accuracy.
Needs Complete typed solution with 100 % accuracy.
A uniform 300 NN trapdoor in a ceiling is hinged at one side. Find the net upward force needed to begin to open it if the upward force is applied at the center. Express your answer with the appropriate units. Find the total force exerted on the door by the hinges if the upward force is applied at the center. Express your answer with the appropriate units Find the net upward force needed to begin to open it if the upward force is applied at the center of the edge opposite the hinges. Express your answer with the appropriate units. Find the total force exerted on the door by the hinges if the upward force is applied at the center of the edge opposite the hinges. Express your answer with the appropriate units.
Problem 1. Two forces act on the screw eye. If F1 = 400 N and F2 = 600 N, determine the angle between them, so that the resultant force has a magnitude of FR = 800 N. рат
A rectangular bar is hinged at point O. The bar is 3.00 m long and 2.00 m wide. Find the torque resulting from the following force about an axis through point O: F1 = 1 N. Use a positive for a counterclockwise torque and a negative for a clockwise torque.
Your answer is partially correct. One end of a wire is attached to a ceiling, and a solid brass ball is tied to the lower end. The tension in the wire is 221 N. What is the radius of the brass ball? Number i 1.75 Units m
Two workers carry a uniform 15.0-ft plank that weighs 22.0lbs. A load of blocks weighing 165 lb is located 7.00 ft from the first worker. What force must each worker exert to hold up the plank and load?
A 3.0kg uniform pole 1.8m long is held horizontal by a hinge to a wall and supported by a cable attached to its center which forms a 45° angle with the pole. A large bird house with a mass of 6.6kg is at the end of tge pole. Find the force that the hinge exerts on the pole.
What is the vertical force, Fy, exerted on the boom by the hinge?