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Physics for scientists and engineers chapter 3 q 50

PostPosted: Thu 05 May, 2011 06:37
by Integr88
Sand moves without slipping at 6.0 m/s down a conveyer that is tilted at 15°. The sand enters a pipe 3.0 m below the end of the conveyer belt, as shown in FIGURE P4.50.
What is the horizontal distance d between the conveyer belt and the pipe?

Re: Physics for scientists and engineers chapter 3 q 50

PostPosted: Thu 05 May, 2011 06:46
by barnamah
Because the sand moves at 15° angle, we have x component and y component. As we want to find the distance between the conveyer belt and the bottom of the pipe which is located 3m bellow the canveyer belt, we need to find the horizontal distance d which is the x distance.
we have:

Initial velocity 6.0m/s
`V_0=6.0m/s`
setting the center of the conveyer belt at 0 in our x and y axis the( please view the image)
x initial `x_i=0m`
x final `x_f`=? we want to find
y initial `y_i=3m`
y final `y_f=0m`
the angle 15° `theta=15°`

1) we find the horizontal velocity `v_(ix)=v*costheta=6.0*cos(15)=5.8m/s`
2)We find vertical velocity `v_(iy)=v*sintheta=6.0*sin(15)=1.6m/s` and velocity due to gravity 8.9m/`s^2` is added to this velocity. Using formula
`y_f=y_i-v_(iy)*Δt-1/2a_yΔt^2` EQ1
`y_f=0m`
`y_i=3m`
`v_(iy)=1.6m/s`
`a_y=-9.8m/s^2`

substitute the above values in EQ1 we have:
`0=3-1.6*Δt-1/2(9.8)Δt^2`
`0=3-1.6*Δt-4.9Δt^2`
This is now a quadratic equation and it is the same as `-4.9x^2-1.6x+3=0` and solving for x. So we solve for t.
Solving for t we get t=-0.96 and t=0.63 out of 2 values the -0.96 is not valid because of the negative time so we use t=0.63s

We can use vertical velocity (velocity in y direction) to find out how long it took to reach to the ground.

`Δt=0.63s` so it takes the sand to reach to the ground in 0.63s. So this is the same time for x direction for the sand to travel the distance d which we call it `x_f`

Using this value we can use `x_f=x_i+v_(ix)*Δt` we can find the `x_f`
`x_f=x_i+v_(ix)*Δt`
`x_f=0+5.8*0.63`
`x_f=3.654m`

`:.` the distance d is 3.654m