Question

Problem 4.49

Sand moves without slipping at 6.0 m/s down a conveyer that is tilted at 15 ?   . The sand enters a pipe h = 3.2m below the end of the conveyer belt, as shown in the figure (Figure 1) .

Part A

What is the horizontal distance d between the conveyer belt and the pipe?

Express your answer to two significant figures and include the appropriate units.

Answer 1

The answer in the end is 3.71m.

Basically, you need to create your equations for both the x-components of the initial velocity (6.0 m/s) and the y-component. So use the general kinematic equations used for situations with a constant acceleration.

so:

d(x-comp) = V(x-comp)*t + 1/2*a(x-comp)*t^2

But, since a(x-comp) = 0 (g is the only acceleration acting on the sand which is along the y-axis)

Therefore, d(x-comp) = V(x-comp)*t

And you know V(x-comp) = 6.0*cos15, so all you need now is to find out is what the time is. You can find the time by using a kinematic formula and using the y-component of velocity since you already have the distance along the y-axis that the sand travels.

So, use d(y-comp) = V(y-comp) + 1/2*a(y-comp)*t^2

so, 3 = 6.0sine15 + 1/2*9.8*t^2

So now you have a quadratic equation and can solve for time

6.0sine15 + 1/2*9.8*t^2 - 3 = 0

So t = 0.63988 and the other value for t was negative so disregard that.

So Therefore, d(x-comp) = V(x-comp)t
= 6.0cos15*0.63988
= 5.80*0.63988
= 3.71 m