March 28, 2011
1. Solve these problems
1. A teacher standing on a 6in by 8in block of wood exerts a pressure of 4 psi. How much does the teacher weigh?
2. A basketball originally has 10 psi of pressure and a volume of 50 mL. If the volume is decreased to 10 mL, what is the resulting pressure?
2. Take out handout from Friday. We will talk about the solutions in 10 minutes.
U3f – L4

A teacher standing on a 6in by 8in block of wood exerts a pressure of 4 psi. How much does the teacher weigh?
Step 1: Write given information
P = 4 psi
F = ??
A = ?
Step 2: Write the formula and solve
A = L*W=6 in *8 in = 48 in 2
P = F/A
4psi = F/48in 2
F = 4psi *48in 2
F=192 lbs
U3f – L4

A basketball originally has 10 psi of pressure and a volume of 50 mL. If the volume is decreased to
10 mL, what is the resulting pressure?
Step 1: Write given information
P1 = 10psi
V1 = 50 mL
P2 = ??
V2 = 10mL
Step 2: Write the formula and solve
P
1
*V
1
=P
2
*V
2
10psi * 50 mL = P
2
*10mL
500 psi*mL = P
2
*10mL
P2 =500 psi*mL/(10mL)
P2=50psi
U3f – L4

A gas system has initial pressure and volume of 80 psi and 6 L. If the volume changes to 10 L, what will the resultant pressure be in psi?
Step 1: Write given information
P1 = 80 psi
V1 = 6 L
P2 = ?
V2 = 10 L
Step 2: Write the formula and solve
P
1
V
1
= P
2
V
2
(80 psi) x (6 L) = P2 x (10 L)
480 psi*L = P2 x (10 L)
10 L 10 L
P2 = 48 psi U3f – L4

A sample of hydrogen gas is compressed from 300 cm 3 to 2.5 cm 3 .
Its pressure is now 5 psi. What was the original pressure of the helium?
Step 1: Write given information
P1 = ?
V1 = 300 cm 3
P2 = 5 psi
V2 = 2.5 cm 3
Step 2: Write the formula and solve
P
1
V
1
= P
2
V
2
P1 x (300 cm 3 ) = 5 psi x (2.5 cm 3 )
P1 x (300 cm 3 ) = 12.5 psi* cm 3
300 cm 3 300 cm 3
P1 = 0.0417 psi U3f – L4

The surface of a concrete block that touches the ground is 0.75 ft 2 .
What is the pressure of the concrete block on the surface of the ground if the block weighs 50 lbs.
Step 1: Write given information
P = ?
A = 0.75 ft 2
F = 50 lbs
Step 2: Write the formula and solve
P=F/A
P=50lbs/0.75
ft 2
P=66.7lbs/ ft 2
U3f – L3

1.
The diagram below illustrates how hydraulic brakes in a car work. The pedal must be pressed with a force of 8 lbs.
The surface area of the piston connected to the pedal is 1 square inch. If the surface area of the piston connected to the other end of the brake line is 1.5 square inch, what is the force applied to that piston?
P1 = F1 / A1
1
2
P2 = F2 / A2
U3f – L3

Step 1: Write given information
P1 = ?
P2 = ?
F1 = 8 lbs.
A1 = 1 in 2
F2 = ?
A2 = 1.5 in 2
Step 2: Write the formula and solve for the unknowns
P1 = F1 / A1
P1 = (8 lbs) / (1 in 2 )
P1 = 8 psi
P1 = P2 = 8 psi
P2 = F2 / A2 F2 = P2 x A2
F2 = 8 psi x 1.5 in 2 = 12 lbs.
U3f – L3

A bicycle tire is holding air at 85 psi of pressure. The valve of the tube lets air in and out. If the cross sectional area of the valve is
0.035 in 2 , how much force is required to add more air to the tube?
Step 1: Write given information
P = 85 psi
A = 0.035 in 2
F = ?
Step 2: Write the formula and solve
P=F/A
85psi=F/0.035
in 2
85psi*0.035
in 2 =F
F =2.975 lbs
U3f – L4

A bicycle pump can be modeled like the diagram in number 7 above. Suppose the tube has 85 psi of pressure inside. To add more air to the tube, how much force is required to push the handle end of a bike pump if its cross-sectional area is 1 in 2
P
1
= F
1
/ A
1
P
1
=P
2
P
2
= F
2
/ A
2
U3f – L4

Step 1: Write given information
P1 = 85psi P2 = 85psi
F1 = 2.975 lbs.
A1 = 0.035 in 2
F2 = ?
A2 = 1 in 2
Step 2: Write the formula and solve for the unknowns
P2 = F2 / A2
85psi= F2 / (1 in 2 )
F2= 85 psi* 1 in 2
F2 = 85 lbs
U3f – L4

If we doubled the amount of force used in number 9 above, what must we change the area of the pump to be in order to maintain the same pressure
Step 1: Write given information
P = 85 psi
A = ???
F = 170 lbs
Step 2: Write the formula and solve
P=F/A
85psi=170lbs/A
85psi*A=170 lbs
A=170lbs/85 psi
A=2in 2
U3f – L4

U3f – L4