Force exerted by the wall?

  • Thread starter DrMcDreamy
  • Start date
  • #1
68
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Homework Statement



This is a two part problem I figured out the 2nd part but not the first.

a) A 7.3 m ladder whose weight is 364 N is placed against a smooth vertical wall. A person whose weight is 810 N stands on the ladder a distance 2.7 m up the ladder. The foot of the ladder rests on the floor 3.577 m from the wall.Calculate the force exerted by the wall. Answer in units of N.

b) Calculate the normal force exerted by the floor on the ladder. Answer in units of N. 1174 N

Homework Equations



Forces:
x-axis: Fk - NWall = 0
y-axis: Nfloor - Wperson - Wladder = 0

Torque:
Nwall x h - Wladder x [tex]\frac{h}{2}[/tex]cos[tex]\theta[/tex] - Wperson x (2.7 x cos[tex]\theta[/tex]) =0

The Attempt at a Solution



[tex]\theta[/tex] = cos-1([tex]\frac{3.577}{7.3}[/tex]) = 61

tan[tex]\theta[/tex]61=[tex]\frac{h}{3.577}[/tex]=6.45

3.65cos61=1.77

2.7 x cos61 = 1.31


Nwall x 6.45 - 364 N x 1.77 - 810 N x 1.31 =0

Nwall x 6.45 -644 - 1061 = 0
Nwall x 6.45 = 1705
Nwall = [tex]\frac{1705}{6.45}[/tex]
Nwall = 264 N

Its wrong. What am I doing wrong?
 

Answers and Replies

  • #2
57
0
Why do you have cosines?

I'm not sure your torque equation under Relevant equations is right. For example, take the cross product of the weight of the ladder with the perpendicular distance between where the force is being applied and the point by which you take the torque. So here, I believe, it would be Mg(d/2), where d is the (horizontal) distance between the point by which you take the origin and the force mg. Or in other words, mg(L/2)sin(theta). You can find theta by using the sides of the triangle.

Remember, torque = r cross f = rfsin(theta)
 
  • #3
57
0
by the way, L = height of ladder.
 
  • #4
68
0
I figured it out! Sorry for my messy work! :tongue:

attachment.php?attachmentid=30348&stc=1&d=1291429570.jpg
 

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