1. The moonpulls on Earth’s oceans and causes the tides. a) Assume the moon is d
ID: 1682149 • Letter: 1
Question
1. The moonpulls on Earth’s oceans and causes the tides.a) Assume the moon is directly overhead. What is the ratio F moon/Fearth of the gravitational forces of the moon and earthexerted on you?
b) How close would the Earth need to be to the Sun (center tocenter distance) in order for the Sun to lift you off of the Earth?Do not guess, calculate your answer!
2. If you place a cubic block of wood of side L=0.1 mand density P wood=600 kg/m^3 on the surface of a lake, the blockwill sink a distance y<L down into thewater.
a) How far will the block sink?
b) A frog of mass m=60 g leaps onto the top of the block. Nowhow far does the block sink?
Hint: draw a free body diagram of the block. Ask yourself: is theblock in equilibrium?
1. The moonpulls on Earth’s oceans and causes the tides.
a) Assume the moon is directly overhead. What is the ratio F moon/Fearth of the gravitational forces of the moon and earthexerted on you?
b) How close would the Earth need to be to the Sun (center tocenter distance) in order for the Sun to lift you off of the Earth?Do not guess, calculate your answer!
2. If you place a cubic block of wood of side L=0.1 mand density P wood=600 kg/m^3 on the surface of a lake, the blockwill sink a distance y<L down into thewater.
a) How far will the block sink?
b) A frog of mass m=60 g leaps onto the top of the block. Nowhow far does the block sink?
Hint: draw a free body diagram of the block. Ask yourself: is theblock in equilibrium?
Explanation / Answer
ME = 5.97 x 1024 kg MS = 1.99 x 1030 kg MM = 7.36 x 1022 kg RE = 6.38 x 106 m = 6.38 Mm DM = 3.84 x 108 m --- The absolute values of the gravitational forces from Earth andMoon are: FE = GmME/RE2 , FM = GmMM/DM2= 0 FM/FE =(GmMM/DM2)/(GmME/RE2)=MMRE2/MEDM2 FM/FE =[(7.36x1022kg)(6.38x106m)2]/[(5.97x1024kg)(3.48x108m)2]= 4.13 x 1022+12-24-16 = 4.13 x10-6 FNET = FE + FS = 0 FNET = -GmME/RE2 +GmMS/DS2 = 0 ME/RE2 =MS/DS2 DS2 =MSRE2/ME =[(1.99x1030kg)(6.38x106m)2]/(5.97x1024kg)= 13.6 x 1030+12-24 m2 = 1.36x 1019 m2 DS = (1.36 x 1019m2)1/2 = 3.69 x 109 m = 3.69Gm --- Density of wood w w = 600 kg/m3 Density of water H2O at 1 atm,20 oC H2O = 998 kg/m3 At equilibrium, upward force of water = downward force of woodblock, so that net force = 0 FNET = Fw + FH2O = -mg +H2OVdispg = 0 where Vdisp = volume of displaced water and the mass of the wood block can be found by: w = mw/vw mw = wvw =(600kg/m3)(0.1m)3 = 0.600 kg FNET = -(0.600kg)[g] +(998kg/m3)Vdisp[g] = 0 (998kg/m3)Vdisp = 0.600kg but Vdisp = X*Y*Z = (0.1m)*Y*(0.1m) =0.01m2Y (998kg/m3)(0.01m2)Y = 0.600kg Y =(0.600kg)/[(998kg/m3)(0.01m2)] = 0.0601 m =6.01 cm If a frog jumps on the wood block, the wood block's massincreases by 60g or 0.060 kg mass of wood block with frog: mF mF = 0.600kg + 0.060kg = 0.660kg FNET = Fw + FH2O =-mFg + H2OVdispg =0 -(0.660kg)[g] + (998kg/m3)(0.01m2)Y[g] =0 Y =(0.660kg)/[(998kg/m3)(0.01m2)] = 0.0661 m =6.61 cm FM = GmMM/DM2= 0 FM/FE =(GmMM/DM2)/(GmME/RE2)=MMRE2/MEDM2 FM/FE =[(7.36x1022kg)(6.38x106m)2]/[(5.97x1024kg)(3.48x108m)2]= 4.13 x 1022+12-24-16 = 4.13 x10-6 FNET = FE + FS = 0 FNET = -GmME/RE2 +GmMS/DS2 = 0 ME/RE2 =MS/DS2 DS2 =MSRE2/ME =[(1.99x1030kg)(6.38x106m)2]/(5.97x1024kg)= 13.6 x 1030+12-24 m2 = 1.36x 1019 m2 DS = (1.36 x 1019m2)1/2 = 3.69 x 109 m = 3.69Gm --- Density of wood w w = 600 kg/m3 Density of water H2O at 1 atm,20 oC H2O = 998 kg/m3 H2O = 998 kg/m3 At equilibrium, upward force of water = downward force of woodblock, so that net force = 0 FNET = Fw + FH2O = -mg +H2OVdispg = 0 where Vdisp = volume of displaced water and the mass of the wood block can be found by: w = mw/vw mw = wvw =(600kg/m3)(0.1m)3 = 0.600 kg FNET = -(0.600kg)[g] +(998kg/m3)Vdisp[g] = 0 (998kg/m3)Vdisp = 0.600kg but Vdisp = X*Y*Z = (0.1m)*Y*(0.1m) =0.01m2Y (998kg/m3)(0.01m2)Y = 0.600kg Y =(0.600kg)/[(998kg/m3)(0.01m2)] = 0.0601 m =6.01 cm If a frog jumps on the wood block, the wood block's massincreases by 60g or 0.060 kg mass of wood block with frog: mF mF = 0.600kg + 0.060kg = 0.660kg FNET = Fw + FH2O =-mFg + H2OVdispg =0 mw = wvw =(600kg/m3)(0.1m)3 = 0.600 kg FNET = -(0.600kg)[g] +(998kg/m3)Vdisp[g] = 0 (998kg/m3)Vdisp = 0.600kg but Vdisp = X*Y*Z = (0.1m)*Y*(0.1m) =0.01m2Y (998kg/m3)(0.01m2)Y = 0.600kg Y =(0.600kg)/[(998kg/m3)(0.01m2)] = 0.0601 m =6.01 cm If a frog jumps on the wood block, the wood block's massincreases by 60g or 0.060 kg mass of wood block with frog: mF mF = 0.600kg + 0.060kg = 0.660kg FNET = Fw + FH2O =-mFg + H2OVdispg =0 -(0.660kg)[g] + (998kg/m3)(0.01m2)Y[g] =0 Y =(0.660kg)/[(998kg/m3)(0.01m2)] = 0.0661 m =6.61 cmRelated Questions
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