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Unformatted text preview: Equation 11.5 gives the force F2 of the output plunger in terms of the force
F1 applied to the input piston as F2 = F1(A2/A1), where A2 and A1 are the corresponding
areas. In this problem the chair begins to rise when the output force just barely exceeds the
weight, so F2 = 2100 N. We are given the input force as 55 N. We seek the ratio of the radii,
so we will express the area of each circular cross section as π r2 when we apply
Equation 11.5.
SOLUTION According to Equation 11.5, we have A2
A1 = F2 or F1 π r22 F2
=
π r12 F1 Solving for the ratio of the radii yields
r2
r1 = F2
F1 = 2100 N
= 6 .2
55 N 35. REASONING We label the input piston as “2” and the output plunger as “1.” When the
bottom surfaces of the input piston and output plunger are at the same level, Equation 11.5,
F2 = F1 ( A2 / A1 ) , applies. However, this equation is not applicable when the bottom surface
of the output plunger is h = 1.50 m above the input piston. In this case we must use Equation 582 FLUIDS 11.4, P2 = P + ρ gh , to account for the difference in heights. In either case, we will see that
1
the input force is less than the combined weight of the output plunger and car.
SOLUTION
a. Using A = π r2 for the circular areas of the piston and plunger, the input force required to
support the 24 500N weight is ( −3 A2 π 7.70 × 10 m
F2 = F1 = ( 24 500 N ) 2
A 1 π ( 0.125 m ) ) 2 = 93.0 N (11.5) b. The pressure P2 at the input piston is related to the pressure P1 at the bottom of the
output plunger by Equation 11.4, P2 = P1 + ρgh, where h is the difference in heights. Setting () () P2 = F2 / A2 = F2 / π r22 , P = F1 / π r12 , and solving for F2, we have
1 π r2 F2 = F1 22 + ρ gh π r22
πr 1 (11.4) () ( −3 π 7.70 × 10 m
= ( 24 500 N ) 2 π ( 0.125 m ) ( )( ) 2 ) ( + 8.30 × 102 kg/m3 9.80 m/s 2 (1.30 m ) π 7.70 × 10−3 m ) 2 = 94.9 N 36. REASONING When a force of magnitude Fp is applied to the piston, the hydraulic fluid
exerts a force of magnitude Fv against the safety valve. The two forces are related by
Equation 11.5: Ap (11.5)
Fp = Fv A v
where Ap is the crosssectional area of the piston and Av is the crosssectional area of the
valve opening. In order for the valve to open, the outward force Fv of the hydraulic fluid
must be at least as great as the inward force Fs exerted on the valve by the compressed
spring: Fv = Fs . The magnitude of the spring force, in turn, is given by Fs = kx
(Equation 10.2 without the minus sign), where k is the spring constant, and x is the amount
by which the spring is compressed. Therefore, we have Chapter 11 Problems Fv = Fs = kx 583 (1) SOLUTION Substituting Equation (1) into Equation 11.5, we obtain Ap Fp = kx A v (2) Both the piston and the valve have circular crosssections, so the area A of either one is
calculated from its radius r according to A = π r 2 . Substituting A = π r 2 into Equation (2),
we obtain the force Fp on the piston needed to open the safety valve:
2 2
π rp 2 rp 0.0281 m = kx = ( 885 N/m )( 0.0100 m ) Fp = kx = 165 N
2
π rv 0.00650 m rv 37. REASONING AND SOLUTION From Pascal's principle, the pressure in the brake fluid at
the master cylinder is equal to the pressure in the brake fluid at the plungers: PC = PP, or
FC
AC = FP
AP or r
π r2
FP = FC
= FC P = FC P
2
r
AC
π rC
C
AP 2 The torque on the pedal is equal to the torque that is applied to the master cylinder so that F l = FC l C or FC = F l
lC Combining the expression for FC with the expression for FP above, we have l
FP = F
lC rp r
C 2 2 0.150 m 1.90 × 10 –2 m = (9.00 N) = 108 N 0.0500 m 9.50 × 10 –3 m 38. REASONING The magnitude F1 of the force the spring exerts on the piston is found from F1 = kx (Equation 10.3, without the minus sign), where k is the spring constant of the
spring, and x is the amount by which the spring is compressed from its unstrained position.
F
The piston and the plunger are at the same height, so the fluid pressure P =
A
(Equation 11.3) at the piston is equal to the fluid pressure at the plunger. Therefore, the 584 FLUIDS A magnitude of the force F2 that the rock exerts on the plunger is given by F2 = F1 2 A 1
(Equation 11.5) where A1 is the area of the piston and A2 is the area of the plunger. The
magnitude F2 of the force the rock exerts on the plunger is equal to the magnitude W = mg
(Equation 4.5) of the rock’s weight, where m is the rock’s mass and g is the magnitude of
the acceleration due to gravity. SOLUTION Solving F1 = kx (Equation 10.3, without the minus sign) for x, we obtain x= F1 (1) k A Solving F2 = F1 2 (Equation 11.5) for F1 and substituting F2 = mg yields
A 1
A A F1 = F2 1 = mg 1 A A 2 2 (2) Substituting Equation (2) into Equation (1), we find that
x= F1 mg A1 ( 40.0 kg ) ( 9.80 m/s 2 ) 15 cm 2 =
= 5.7 × 10 −2 m = 2 k
k A2 1600 N/m 65 cm 39. SSM REASONING The pressure P ′ exerted on the bed of the truck by the plunger is
P ′ = P − Patm . According to Equation 11.3, F = P ′A , so the force exerted on the bed of the c h truck can...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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