Simple Solutions for energy exercises

PHY 111 COLLEGE PHYSICS ICLASS ASSIGNMENT SCHEDULE: Sections 12147/49 – 12148/50 Fall 2011

9.3 Four acrobats of mass75.0 kg, 68.0 kg, 62.0 kg and 55.0kg for a human tower, with each acrobat

standing on the shoulders of another acrobat. The 75.0kg acrobas is at the bottom of the tower. (a)

What is the normal force acting on the 75—kg acrobat? (b) If the area of each of the 75.0 kg

acrobat’s show is 425 cm2, what average pressure ( not including atmospheric pressure) does the

column of acrobats exert on the floor? (c) Will the pressure be the same if a different acrobat is on

the bottom?

Instructor: Dr. Aquair Muhammad [BABA] Office: Room 7425; Phone: 488-4700 Ext 2436 e-mail: [email protected] Office hours: Monday/Wednesday 4:00pm-6:00pm, Tuesday/Thurs 8:00am-10:00am

November 02, 2011 28 Chapter 9 3,4,11, 23, 31 47

Weight of the four Acrobats is

= A1 + A2 +A3 +A4 = -2548N ( Earthward )

A4= 75.0kg x 9.8 m/s2=735 N

A3= 68.0kg x 9.8 m/s2=666 N

A2 = 62.0kg x 9.8 m/s2=608 N

A 1= 55.0kg x 9.8 m/s2=539 N

A) The Normal

force is equal

and opposite to the force of gravity and is perpendicular to the surface of the

floor. = 2548

B) Average Pressure P on the floor

P = Force/Area

P = W/ A

P = 2548 N/2(.04M)

P = 31.8 x 103 Pa

C) If the area of either of the feet of another acrobats is different than that of the 75.0kg acrobat, then the pressure will also be different. If the area of the feet of another acrobat is the same are the 75.0kg acrobat, then the pressure will be the same.

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9.4 Calculate the mass of a solid gold rectangular bar that has dimensions of 4.5 cm x 11.0 cm x 26.0 cm.



ρgold = 19.3 x 103 kg/M3

Density = mass/volume, or ρ= m/v solve for mass m yields,

m= ρv. In this case ρ of gold is given and volume

V = (4.5 x 11.0 x 26.0) cm3 = (.045 x .11 x .26) m3 =1.29 x 10-3 M3

m=(19.3 x 103 kg/M3) x (1.29 x 10 -3 M) =19.47 kg

m = 19.47kg



9.11 A plank 2.00cm thick and 15.0cm wide is firmly attached to the railing of a ship by clamps so that

the rest of the board extends 2.00 m horizontally over the sea below. A man of mass 80.00 kg is

forces to stand on the very end. If the end of the board drops by 5.00cm because of the man’s

weight, find the shear modulus of the wood

Two cross-sectional areas in the plank, with one directly above the rail and one at the outer end of the plank,

separated by distance and each with area, move a

distance parallel to each other. The force causing this shearing effect in the plank is

the weight of the man applied perpendicular to the length of the plank at its outer end. Since the

shear modulus S is given by

we have



2.00cm x 15.0cm x 2.00 m

F=80.00kgx 9.8 m/s2



9.23 A collapsible plastic bas contains a clusoe solution. If the average gauge

pressure in the vein is 1.33 x 103 Pa, whatmustbe the minimum height h

of the bag in ore vein? Assume the specific gravity of the solution is 1.2.

The density of the solution is . The gauge

pressure of the fluid at the level of the needle must equal the gauge pressure

in the vein, so , and





9.31 A small ferryboat is 4.00m wide and 6.00m long. When a loaded small truck pulls onto it., the boat

sinks an additional 4.00cm into the river. What is the weight of the truck?

The boat sinks until the weight of the additional water displaced equals the weight of the truck. Thus,

or





9.47A hypodermic syringe contains a medicine with the density of

water. The barrel of the syringe has a cross-sectional area

of 2.5 x 10 -5 m2 . In the absence of a force on the plunder,

the pressure every where is and 1.00atm. a force F of

magnitude 2.00 N is exerted on the plunder, making

medicine squirt from the needle. Determine the medicine’s

flow speed through the needle. Assume the pressure in the

needle remains equal to 1.00 atm that the syringe is horizontal.

From Bernoulli’s equation, choosing at y = 0 the level of the syringe and needle, so the flow speed in the needle,

is

In this situation,

Thus, assuming 0,




Simple Solutions for energy exercises

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