solutions to chapter 12 pt 1a

Prévia do material em texto

�PAGE �912�
 	Chapter 12 
 Static Equilibrium and Elasticity
�PAGE �913�
Static Equilibrium and Elasticity
 
Chapter 12
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
17	•	
	Picture the Problem Choose a coordinate system in which upward is the positive y direction and to the right is the positive x direction and use the conditions for translational equilibrium.
	(a) Apply 
 to the forces acting on the tip of the crutch:
	
 (1)
and
 (2) 
	Solve equation (2) for Fn and assuming that fs = fs,max, obtain:
	
	Substitute in equation (1) and solve for µs:
	
	(b) 
	(c) 
The Center of Gravity
18	•	
	Picture the Problem Let the weight of the automobile be w. Choose a coordinate system in which the origin is at the point of contact of the front wheels with the ground and the positive x axis includes the point of contact of the rear wheels with the ground. Apply the definition of the center of gravity to find its location.
	Use the definition of the center of gravity:
	
or, because W = w, 
	Solve for xcg:
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
Some Examples of Static Equilibrium
22	•	
	Picture the Problem We can use the given definition of the mechanical advantage of a lever and the condition for rotational equilibrium to show that M = x/X.
	(a) Express the definition of mechanical advantage for a lever:
	
	Apply the condition for rotational equilibrium to the lever:
	
	Solve for the ratio of F to f to obtain:
	
	Substitute to obtain:
	
	(b) 
23	•
	Picture the Problem The force diagram shows the tension in the forestay, 
 the tension in the backstay, 
 the gravitational 
force on the mast
 and the force exerted by the deck, 
 Let the origin of the coordinate system be at the foot of the mast with the positive x direction to the right and the positive y direction upward. Because the mast is in equilibrium, we can apply the conditions for both translational and rotational equilibrium to find the tension in the backstay and the force that the deck exerts on the mast. 
	
	Apply 
to the mast about an axis through its foot and solve for TB:
	
and
	Find (F, the angle of the forestay with the vertical:
	
	Substitute to obtain:
	
	Apply the condition for translational equilibrium in the x direction to the mast:
	
or
	Apply the condition for translational equilibrium in the y direction to the mast:
	
or
	Because FDcos( = 0:
	
, 
and
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
*27	•	
	Picture the Problem The planes are frictionless; therefore, the force exerted by each plane must be perpendicular to that plane. Let 
be the force exerted by the 30( plane, and let 
 be the force exerted by the 60( plane. Choose a coordinate system in which the positive x direction is to the right and the positive y direction is upward. Because the cylinder is in equilibrium, we can use the conditions for translational equilibrium to find the magnitudes of 
and 
.
	
	Apply 
 to the cylinder:
	
 (1) 
	Apply 
 to the cylinder:
	
 (2)
	Solve equation (1) for F1:
	
 (3)
	Substitute in equation (2) to obtain:
	
	Solve for F2:
	
or
	Substitute in equation (3):
	
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32	••	
	Picture the Problem The diagram shows the forces 
and
acting at the supports, the weight of the board 
 acting at its center of gravity, and the weight of the diver 
acting at the end of the diving board. Because the board is in equilibrium, we can apply the condition for rotational equilibrium to find the forces at the supports. 
	
	Apply
about an axis through the left support:
	
	Solve for F2:
	
	Substitute numerical values and evaluate F2:
	
	Apply
about an axis through the right support:
	
	Solve for F1:
	
	Substitute numerical values and evaluate F1:Ladder Problems
*48	••	 
	Picture the Problem The ladder and the forces acting on it at the critical moment of slipping are shown in the diagram. Use the coordinate system shown. Because the ladder is in equilibrium, we can apply the conditions for translational and rotational equilibrium.
	
	Using its definition, express µs:
	
 (1)
	Apply 
about the bottom of the ladder:
	
	Solve for FW:
	
	Find the angle (:
	
	Evaluate FW:
	
	Apply 
to the ladder and solve for fs,max:
	
and
	Apply 
to the ladder:
	
	Solve for Fn:
	
	Substitute numerical values in equation (1) and evaluate µs:
	
49	••	
	Picture the Problem The ladder and the forces acting on it are shown in the diagram. Because the wall is smooth, the force the wall exerts on the ladder must be horizontal. Because the ladder is in equilibrium, we can apply the conditions for translational and rotational equilibrium to it.
	
	Apply 
to the ladder and solve for Fn:
	
( 
	Apply 
to the ladder and solve for fs,max:
	
( 
	Apply 
about the bottom of 
the ladder:
	
	Solve for x:
	
	Referring to the figure, relate x to h and solve for h:
	
and
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
Stress and Strain
*54	•	
	Picture the Problem L is the unstretched length of the wire, F is the force acting on it, and A is its cross-sectional area. The stretch in the wire (L is related to Young’s modulus by 
 We can use Table 12-1 to find the numerical value of Young’s modulus for steel.
	Find the amount the wire is stretched from Young’s modulus:
	
	Solve for (L:
	
	Substitute for F and A to obtain:
	
	Substitute numerical values and evaluate (L:
	
55	•	
	Picture the Problem L is the unstretched length of the wire, F is the force acting on it, and A is its cross-sectional area. The stretch in the wire (L is related to Young’s modulus by 
 
	(a) Express the maximum load in terms of the wire’s breaking stress:
	
	Substitute numerical values and evaluate Fmax:
	
	(b) Using the definition of Young’s modulus, express the fractional change in length of the copper wire:
	
56	•	
	Picture the Problem L is the unstretched length of the wire, F is the force acting on it, and A is its cross-sectional area. The stretch in the wire (L is related to Young’s modulus by 
 We can use Table 12-1 to find the numerical value of Young’s modulus for steel.
	Find the amount the wire is stretched from Young’s modulus:
	
	Solve for (L:
	
	Substitute for F and A to obtain:
	
	Substitute numerical values and evaluate (L:
	
*57	•	
	Picture the Problem The shear stress, defined as the ratio of the shearing force to the area over which it is applied, is related to the shear strain through the definition of the shear modulus; 
. 
	Using the definition of shear modulus, relate the angle of shear, ( to the shear force and shear modulus:
	
	Solve for ( :
	
	Substitute numerical values and evaluate ( :
	
58	••	
	Picture the Problem The stretch in the wire (L is related to Young’s modulus by
, where L is the unstretched length of the wire, F is the force acting on it, and A is its cross-sectional area. For a composite wire, the length under stress is the unstressed length plus the sum of the elongations of the components of the wire.
	Express the length of the composite wire when it is supporting a mass of 
5 kg:
	
 (1)
	Express the change in length of the composite wire:
	
	Find the stress in each wire:
	
	Substitute numerical values and evaluate (L:
	
	Substitute in equation (1) and evaluate L:
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
72	•	
	Picture the Problem We can determine the ratio of L to h by noting the number of ropes supporting the load whose mass is M.
	(a) Noting that three ropes support the pulley to which the object whose mass is M is fastened we can conclude that:
	
	(b) Apply the work-energy principle to the block-tackle object to obtain:
	
or�
�
�
�Note that these answers are not those of the 4e.
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n
Offset
Offset as a function of n
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