Stephen is pushing his sister Joyce in a wheelbarrow when itis stopped by a bric

Question

Stephen is pushing his sister Joyce in a wheelbarrow when itis stopped by a brick 7.83 cm high.The handles make an angle of 15.0° below the horizontal. Adownward force of 402 N is exerted onthe wheel, which has a radius of 19.1cm. (a) What force must Stephen apply along thehandles in order to just start the wheel over the brick?
1 N

(b) What is the force (magnitude and direction) that the brickexerts on the wheel just as the wheel begins to lift over thebrick? Assume in both parts that the brick remains fixed and doesnot slide along the ground.
Magnitude
2 kN
Direction
3° above the horizontal (to theleft)
Stephen is pushing his sister Joyce in a wheelbarrow when itis stopped by a brick 7.83 cm high.The handles make an angle of 15.0° below the horizontal. Adownward force of 402 N is exerted onthe wheel, which has a radius of 19.1cm. (a) What force must Stephen apply along thehandles in order to just start the wheel over the brick?
1 N

(b) What is the force (magnitude and direction) that the brickexerts on the wheel just as the wheel begins to lift over thebrick? Assume in both parts that the brick remains fixed and doesnot slide along the ground.
Magnitude
2 kN
Direction
3° above the horizontal (to theleft)
(a) What force must Stephen apply along thehandles in order to just start the wheel over the brick?
1 N

(b) What is the force (magnitude and direction) that the brickexerts on the wheel just as the wheel begins to lift over thebrick? Assume in both parts that the brick remains fixed and doesnot slide along the ground.
Magnitude
2 kN
Direction
3° above the horizontal (to theleft)

Explanation / Answer

   from the figure in the text book the forces are
   Fx = F cos15.0o
   Fy = F sin15.0o
   as the wheel leaves the ground the ground exerts noforce on it so we get
   Fx = 0
    F cos15.0o - nx = 0........ (1)
   Fy = 0
    F sin15.0o - 402 N +ny = 0 ........ (2)
   as we take the torques about the contact point withthe brick
   so the needed distances will be
   b = R - 7.83 cm
      = (19.1 - 7.83) cm
      = 11.27 cm
   a = (R2 - b2)
      = ........ cm
(a)
   as = 0
   - Fx b + Fy a + (402 N) a =0
   F = ........ N
(b)
   using the equations (1) and (2) we get
   nx = F cos15.0o
       = ........ N amd
   ny =  (402 N)+ F sin15.0o
       = ........ N
   n = (nx2 +ny2)
      = ........ N
   = tan-1(ny /nx)
      = ........o towards theleft and upward

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