Static Equilibrium and Fluids

Question

1. A street light is supported by two wires and each makes a different angle from each other. The one on the left is 32 degrees and the one on the right is 43 degrees. Find the tension in each wire if the street light has a mass of 60.0 kg.
2. There are three masses m1 = 2.0 kg, m2 = 5.0 kg and m3 = 7.0 kg. The mass are all affixed to a massless rod of length 15cm. m1 and m3 are at the ends and m2 is 10 cm from left where m1 is. Calculate the center of mass of the system.
3. A pair of hedge trimmers are 45 cm long from the pivot to the end of the handles. If the person applies of a force of 15 N on each handle and the blades are 3.5 cm long, how much force is applied to the branch?
4. A 245 kg person lays on board that is supported only on the ends and that has a mass of 12.0 kg. If the person and the board are both 205 cm long and the person%u2019s center of mass is 38 cm from the left side, what is the force on either side where the board is supported?
5. A large tank of water is full and has a depth of 10.0 m. What is the pressure at the bottom of the tank? What would the velocity of the water be if the tank was opened at the bottom?
6. A Two balloons of identical volume (radius 11.5 m) are filled with gas. The first with helium, the second with hydrogen (H2) gas. Each balloon and cage have a mass of 750 kg. What is the lift created by each one? Which one lifts better?
7. A irregular piece of metal has a dry mass of 5.7 kg. Once it is submerged it has an apparent mass of 2.4 kg. What is the density of the piece of metal?
8. If water leaves a tank of water at a velocity of 16.7 m/s, how tall is the tank?

Explanation / Answer

1)Let Tl be the tension in the left wire,

Tr be the tension in right wire,

then by force balance we have

Tl cos 32+Tr cos 43 = mg = 60*9.8 = 588 N

0.848Tl+0.731Tr = 588.

and in the x direction, Tl sin 32 = Tr sin 43

0.53Tl = 0.682 Tr

solving, Tl = 415.23 N

Tr = 322.68 N

2)centre of mass = m1x1+m2x2+m3x3/(m1+m2+m3) = (2*0+5*10+7*15)/(2+5+7) = 11.07 cm

3)The torque has to be zero for each of the handle,

then at the end of blade, for each trimmer let the force be F

then for making the torque zero on the whole system

F*3.5 = 15*45

F= 192.857 N

then force applied to the branch = 2*F= 2*192.857 = 385.714 N

4)Let the force from the support on left side be Fl,

force from the support on right side be Fr

then it can be considered that the mass of the person acts downwards from his centre of mass.

the Fm = mg= 245*9.8 = 2401 N at x = 0.38 m. from left.

the mass of the board acts at middle, it's centre of mass will be at x= 205/2 = 102.5 cm = 1.025 m

then Fb = 12*9.8 = 117.6 N at x= 1.025 m

then by force balance, Fl+Fr = Fb+Fm

Fl+Fr = 2518.6

by torque balance with respect to point x=0.

Fm*0.38+Fb*1.025 = Fr*2.05

2401*0.38 +117.6*1.025 = Fr*2.05

Fr = 503.863 N.

Fl = 2518.6-503.863 = 2014.737 N.

both are upwards.

5)pressure = rho*g*h = 1000*9.8*10 = 98000 Pa = 98000 N/m^2

velocity = sqrt((2*P)/(rho) )= sqrt((2*98000)/1000)) = 14 m/s

6)density of air = 1.293 Kg/m^3

volume of ballon = 1.33*pi*r^3 = 1.333*pi*(11.5^3)= 6369.03 m^3

lift created by helium ballon :

buoyant force = rho air )*V*g= 1.293*6369.03 *9.8= 80704.53 N

density of Helium= 0.1785 Kg/m^3

mass of gas = volume of ballon*density of helium = 0.1785 *6369.03 = 1136.87 Kg

mass of the ballon+cage+gas = 1136.87+750 = 1886.87 Kg.

weight of total system=1886.87*9.8 = 18491.326 N

lift = buoyant force - weight = 80704.53-18491.326 = 62213.204 N

lift created by H2 ballon :

buoyant force = rho air )*V*g= 1.293*6369.03 *9.8= 80704.53 N

density of H2= 0.089 Kg/m^3

mass of gas = volume of ballon*density of helium = 0.089 *6369.03 = 566.84367Kg

mass of the ballon+cage+gas = 566.84367+750 =1316.84367 Kg.

weight of total system=1316.84367*9.8 = 12905.067 N

lift = buoyant force - weight = 80704.53-12905.067=67799.463 N

7)apparent mass= dry mass-(volume*density of water)

2.4 = 5.7-(volume*1000)

volume = dry mass/density = 5.7/density

2.4= 5.7-(5.7*1000)/density

calculating density of solid = 1727.27 Kg/m^3.

8)veolcity = sqrt(2P/rho) = 16.7

16.7^2 = 2P/1000

P = 139445 Pa.

height = P/(rho*g) = 139445/(9.8*1000)= 14.23 m.

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