quiz samples for chapter 12 general physics i may 11, 2020...

General Physics IQuiz Samples for Chapter 12Equilibrium and Elasticity

May 11, 2020

Name: Department: Student ID #:

Notice

� +2 (−1) points per correct (incorrect) answer.

� No penalty for an unanswered question.

� Fill the blank ( ) with � (8) if the statement iscorrect (incorrect).

� Textbook: Walker, Halliday, Resnick, Principlesof Physics, Tenth Edition, John Wiley & Sons(2014).

12-1 Equilibrium

1. A rigid body at rest is said to be in staticequilibrium.

(a) (�) For such a body, the vector sum of theexternal forces acting on it is zero:

Fnet = 0.

(b) (�) If all the forces lie in the xy plane, thisvector equation is equivalent to twocomponent equations:

Fnet,x = 0, Fnet,y = 0.

(c) (�) Static equilibrium also implies that thevector sum of the external torques acting onthe body about any point is zero, or

τnet = 0.

(d) (�) If the forces lie in the xy plane, all torquevectors are parallel to the z axis, and thebalance-of-torques equation is equivalent tothe single component equation:

τnet,z = 0.

(e) (�) If the gravitational acceleration is thesame for all the elements of the body, thecenter of gravity is at the center of mass.

2. (�) For an object in equilibrium the sum of thetorques acting on it vanishes only if each torque iscalculated about a single point.

3. (�) For a body to be equilibrium under thecombined action of several forces: the sum of thecomponents of all the forces in any direction mustequal zero.

4. (�) To determine if a rigid body is in equilibriumthe vector sum of the gravitational forces acting onthe particles of the body can be replaced by asingle force acting at the center of gravity.

12-2 Some Examples of Static Equilibrium

1. The following figure shows a box of mass Mhanging by a rope of negligible mass and tensionTy from a boom that consists of a uniform hingedbeam of mass m and horizontal cable of length b,negligible mass, and tension Tx. The heightdifference between the hinge and the cable is a.

Cable

Hinge

Beam comRope

The beam makes an angle of θ with the horizontalx axis:

tan θ =a

b.

The hinge can exert a force with nonvanishingcomponents along x and y directions.

F = Fx i + Fy j.

We place the origin O of the xy plane at the hinge.

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May 11, 2020

(a) (�) We employ the following identifiers:1 =hinge, 2 =beam weight, 3 =cable tension,4 =rope tension. All of the forces applied tothe beam are

F1 = Fx i + Fy j,

F2 = −mgj,F3 = −Tx i,F4 = −Ty j = −Mgj.

(b) (�) The sum of the forces is the net force onthe beam:

Fnet = (Fx − Tx)i + (Fy −mg −Mg)j.

(c) (�) All of the torques applied to the beam,calculated about the origin, are

τ1 = 0× (Fx i + Fy j) = 0,

τ2 =

(b

2i +

a

2j

)× (−mgj) = − b

2mgk,

τ3 = (bi + aj) × (−Tx i) = aTxk,

τ4 = (bi + aj) × (−Ty j) = −bMgk.

(d) (�) The sum of the torques is the net torqueon the beam:

τnet =

(aTx − bMg − 1

2mgb

)k.

(e) (�) The solutions are

Tx =b

2a(m+ 2M)g,

Ty = Mg,

Fx =b

2a(m+ 2M)g,

Fy = (m+M)g.

2. A picture P of weight W is hung by two strings asshown. The magnitude of the tension force of eachstring is T .

(a) (�) The forces are

left string : F1 = −iT cos θ + jT sin θ,

right string : F2 = +iT cos θ + jT sin θ,

P : F3 = −W j.

(b) (�) The net force is

Fnet = j(2T sin θ −W ).

(c) (�) Therefore,

W = 2T sin θ.

3. A uniform plank (ignore its thickness) of length Land mass M is supported by two normal forces FA

and FB at the end points A and B, as shown.

We set the coordinate system so that A and B areat the origin and at Li, respectively. The supportat A is then moved to X, whose position is xi. Thesupporting forces at A and B are then:

(a) (�) The forces applied to the plank are givenby

FA = FAj,

FB = FB j,

Fg = −Mgj,

where Fg is the gravitational force acting onthe center of gravity.



May 11, 2020

(b) (�) At the initial stage, the torques applied tothe plank, calculated about the origin, aregiven by

τA = xi× FAj = FAxk,

τB = Li× FB j = FBLk,

τg =L

2i× (−Mgj) = −1

2MgLk,

where Fg is the gravitational force acting onthe center of gravity.

(c) (�) The net force and net torque are given by

Fnet = (FA + FB −Mg)j,

τnet =

(FAx+ FBL− 1

2MgL

)k.

(d) (�) Because both the net force and torquemust vanish, FA and FB are determined as

FA =MgL

2(L− x)j,

FB =Mg(L− 2x)

2(L− x)j.

(e) (�) If x >L

2, then FB is parallel to −j. This

is unphysical because the normal force mustbe parallel to j. Thus it is impossible to move

A for x >L

2. If we move to a position with

x >L

2, then the equilibrium breaks down.

12-3 Elasticity

1. Three elastic moduli are used to describe theelastic behavior (deformations) of objects as theyrespond to forces that act on them.

(a) (�) Stress is the force per unit area on a bodythat tends to cause it to change shape:

stress =F

A,

where F is the deforming force and A is thecross sectional area of the body that acceptsthe force.

(b) (�) The strain (fractional change in length) islinearly related to the applied stress (force perunit area) by the proper modulus, accordingto the general relation

stress = modulus × strain.

(c) (�) Tensile stress is caused by pulling abody. The applied force tends to elongate thebody. The cross section is perpendicular tothe force. Like the tension, two forces are on asingle line. The deformation (elongation) ∆Lis along the direction of the force.

tensile stress =F

A,

tensile strain =∆L

L,

F

A= E

∆L

L,

where E is the Young’s modulus for theobject.

(d) (�) Shearing stress is caused by a torque.While the forces are coplanar with the crosssection, the torque is perpendicular to thecross section. The deformation ∆x is alongthe direction of the force. If the initial crosssection is a rectangle, then the deformed crosssection is a parallelogram.



May 11, 2020

shearing stress =F

A,

shearing strain =∆x

L,

F

A= G

∆x

L,

where G is the shear modulus of the object.

(e) (�) A solid sphere subject to uniformhydraulic stress from a fluid shrinks involume by an amount ∆V .

hydraulic compression = p =F

A,

hydraulic strain = −∆V

V,

p = −B∆V

V,

where B is the bulk modulus of the object.

2. (�) Stress and pressure have common physicaldimensions.

3. (�) Young’s modulus, shear modulus, bulkmodulus, and pressure have common physicaldimensions.

4. Consider a stress versus strain plot.

Strain ( )

Stress ( ) rupture

(a) (�) For a substantial range of appliedstresses, the stress–strain relation is linear andthe specimen recovers its original dimensionswhen the stress is removed.

(�) In the linear region, modulus is aconstant.

(b) (�) Young’s modulus can be used to calculatethe strain for a stress that is well below theyield strength Sy.

(c) (�) If the stress is increased beyond the yieldstrength Sy of the specimen, the specimenbecomes permanently deformed.

(d) (�) If the stress continues to increase, thespecimen eventually ruptures, at a stresscalled the ultimate strength Su.


quiz samples for chapter 12 general physics i may 11, 2020...

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