Problems and Solution On Elasticity
Problems and Solution On Elasticity
Problems and Solution On Elasticity
If 30% of the
weight of a woman weighing 480 N is supported by each heel, find the stress on each heel
F
Solution: Stress = , where F = 0.30 ( weight ) and A = π r 2
A
0.30 ( 480 N )
Thus, Stress = = 1.8 × 106 Pa
π ( 5.0 × 10 m )
−3 2
Problem 2: For safety in climbing, a mountaineer uses a nylon rope that is 50 m long and 1.0 cm in
diameter. When supporting a 90-kg climber, the rope elongates 1.6 m. Find its Young’s modulus.
F L0 π d2
Solution: Using Y = with A = and F = mg , we get
A ( ΔL) 4
4 ⎡⎣( 90 kg )( 9.80 m s 2 ) ⎤⎦ ( 50 m )
Y= = 3.5 × 108 Pa
π ( 1.0 × 10 m ) ( 1.6 m )
−2 2
Problem 3: A stainless steel orthodontic wire is applied to a tooth, as in the figure below. The wire has an
unstretched length of 3.1 cm and a diameter of 0.22 mm. If the wire is stretched 0.10 mm, find the
magnitude and direction of the force on the tooth. Disregard the width of the tooth and assume the Young’s
modulus for stainless steel is 18 x 1010 Pa.
F L0
Solution: From Y = , the tension needed to stretch the wire
A ( ΔL) 30° 30°
by 0.10 mm is F F
Y A ( ΔL) Y (π d ) ( ΔL)
2
F= =
L0 4 L0
= = 22 N
4 ( 3.1 × 10-2 m )
The tension in the wire exerts a force of magnitude F on the tooth in each direction
along the length of the wire as shown in the above sketch. The resultant force exerted
on the tooth has an x-component of Rx = ΣFx = − Fcos 30° + Fcos 30° = 0 , and a y-
component of Ry = ΣFy = − F sin 30° − F sin 30° = − F = −22 N .
Solution: (a) When at rest, the tension in the cable equals the weight of the 800-kg
F L0
mass, 7.84 × 103 N . Thus, from Y = , the initial elongation of the
A ( ΔL)
cable is
ΔL =
F ⋅ L0
=
( 7.48 × 10 N )( 25.0 m ) = 2.45 × 10−3 m= 2.5 mm
3
(b) When the load is accelerating upward, Newton’s second law gives
F − mg = may , or F = m ( g + ay ) (1)
(c) From the definition of the tensile stress, stress = F A , the maximum
tension the cable can withstand is
Then, equation (1) above gives the mass of the maximum load as
Fmax 8.8 × 10 4 N
mmax = = = 6.9 × 103 kg .
g + a ( 9.8+3.0 ) m s 2
Problem 5: Bone has a Young’s modulus of about 18 x 109 Pa. Under compression, it can withstand a
stress of about 160 x 106 Pa before breaking. Assume that a femur (thigh bone) is 0.50 m long and calculate
the amount of compression this bone can withstand before breaking
⎛ F ⎞⎛ L ⎞ ⎛L ⎞
Solution: From Y = ⎜ ⎟⎜ 0 ⎟ = ( stress ) ⎜ 0 ⎟ , the maximum compression the femur
⎝ A ⎠⎝ ΔL ⎠ ⎝ ΔL ⎠
can withstand is
ΔL =
( stress ) ( L0 )
=
(160 × 10 6
Pa )( 0.50 m )
= 4.4 × 10−3 m = 4.4 mm .
Y 18 × 109 Pa