PHY-133 Section 60 Analytical Physics I Fall 2017 6) A bunch of fake Indians are

Question

PHY-133 Section 60 Analytical Physics I Fall 2017 6) A bunch of fake Indians are on a ship docked in Boston Harbor and want to push all of tea over the edge of the ship, into the water. They set up a ramp to do it, as shown The of the ship is 15.0 m above the surface of the water. Each crate of tea has a mass They push the crates of tea all the way u the 10.0m ramp, two at a time, starting bottom. The static coefficient of friction between a crate and the ramp is 0.350 andt coefficient of friction between a crate and the ramp is 0.250. long, and its end in Fig.-6. is propped up so that it is 5.00 m above the deck. The deck b of 50.0 kg from the he kinetic d. a) If the Indians push with a combined force, F 1, 100 N parallel to the ramp, wha horizontal distance from the edge of the ramp, d, that the crates hit th b) On one o t their runs up the ramp, the top crate falls off half way up the ramp. The In dians continue to push the remaining crate the rest of the way up the ramp with the same combined force of F 1, 100 N parallel to the ramp. What is the horizontal distance from the edge of the plank, d, that the single remaining crate now hits the water? On another run up the ramp with a pair of crates, half the Indians fall off the ramp half way up, and drown. Without pausing, the remaining Indians push the two crates the rest of the way up the ramp. Now what is the horizontal distance from the edge of the ramp d, that the crates hit the water? (Assume that the remaining Indians can push with half the force, or 550 N.) c) d) The remaining Indians now push another pair of crates up the ramp. Now what is the horizontal distance from the edge of the ramp, d, that the crates hit the water? 10,0 m 5.00 m 15.0 m Figure 6 - (a) Indians push pairs of 50.0kg crates up a 10.0 m ramp with a force, P to the ramp. The right side of the ramp is 5.00m above the deck of the ship, and 20.0m above the water surface.

Explanation / Answer

given length of ramp, l = 10 m

h = 5 m

H = 15 m

kinetic coefficient of friciton k = 0.25

static coefifcient of friciotn K = 0.35

m = 50 kg

a. F = 1100 N

so acceleration along the incline, a = F/2m - gsin(theta) - kgcos(theta)

here sin(theta) = 5/10 = 0.5

so theta = 30 deg

hence

a = 1100/100 - g(sin(30) + 0.25*cos(30)) = 3.971 m/s/s

so speed just before the block moves off the ramp = v

2*a*l = v^2

v = sqroot(2*3.971*10) = 8.911 m/s

now this becomes a projecftile problem

time of flight = t

-(H + h) = vsin(theta)*t - 0.5*g*t^2

4.9t^2 - 4.455*t - 20 = 0

solving for t

t = 2.522 s

hence horizontal distance d = vcos(theta)*t = 19.469 m

b. when the create falls half way

initial acceleration a1 = 3.971 m/s/s

hence speed at this point u = sqroot(2*3.971*5) = 6.301 m/s

final acceleration a2 = 1100/50 - g(sin(30) + 0.25*cos(30)) = 14.971 m/s/s

final speed = v

v^2 - u^2 = 2*14.971*5

v = 13.762 m

time of flight = t

-(H + h) = vsin(theta)*t - 0.5*g*t^2

4.9t^2 - 6.8813*t - 20 = 0

solving for t

t = 2.838 s

hence horizontal distance d = vcos(theta)*t = 33.825 m

c. when the indians falls half way

initial acceleration a1 = 3.971 m/s/s

hence speed at this point u = sqroot(2*3.971*5) = 6.301 m/s

final acceleration a2 = 550/100 - g(sin(30) + 0.25*cos(30)) = -1.5289 m/s/s

final speed = v

v^2 - u^2 = 2*(-1.5289)*5

v = 4.94 m/2

time of flight = t

-(H + h) = vsin(theta)*t - 0.5*g*t^2

4.9t^2 - 2.47*t - 20 = 0

solving for t

t = 2.285 s

hence horizontal distance d = vcos(theta)*t = 9.7785 m

d. the remaining indians cannot push the crate up the ramp, as force required to push the crates up the ramps is

F = 2mgsin(theta) + 2kmgcos(theta) = 702 N > 550 N that the indfians can apply

hence the block will not move

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