What is the horizontal distance between a tilted conveyer belt and a pipe?

  • Thread starter Thread starter burton95
  • Start date Start date
  • Tags Tags
    Incline
Join the discussion
Registration is free. Start your own thread to ask a follow-up.
Sign up Log in
7 replies · 2K views
burton95
Messages
54
Reaction score
0

Homework Statement



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

Xi = 0 m
Ti = 0 s
Viy = 6.0 sin θ
Yi = 3 m
Vix = 6.0 cos theta
Xf = ?
Tf = t
Vf =
Yf = 0

Homework Equations



Yf = Yi + Viy(t) - 1/2g(t)2
Xf = Xi + Vix(t)

The Attempt at a Solution



3 + 6.0sin15t -4.9t2

t= .956797

Xf = 0 + 6*cos15*.956797 = 5.54 meters

not it
 
Physics news on Phys.org

Discussion

burton95 said:
3 + 6.0sin15t -4.9t2

Did you mean to write an equation here? What is the sign of the initial y-component of velocity?
 
Hi.

Yf = Yi + Viy(t) - 1/2g(t)2

0 = 3 + 6.0sin15t -4.9t2

the sign on initial velocity is +
 
burton95 said:
the sign on initial velocity is +

The problem says that the sand moves down the conveyor. So, the y-component of velocity should be down.
 
I thought that the direction was accounted for in setting Yi = 3 and Yf = 0.

so my proper equation 0 = 3 - 6.0sin15t -4.9t2
 
That looks good. Note that you took yf =0 and yi = 3 which is correct for taking upward as positive. You also used -9.8 m/s^2 for the acceleration which agrees with taking upward as positive since the acceleration is downward. Likewise, the y component of the initial velocity is downward, so it should be negative when taking upward as positive.
 
Is this correct
Yi = Yo + Voy (Tf - Ti) - (1/2)(g)(Tf-Ti)^2

Because using -g turns my squared term positive