What Is the Force Exerted by the Wall on a Ladder?

[DrMcDreamy]

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: F_k - N_Wall = 0
y-axis: N_floor - W_person - W_ladder = 0

Torque:
N_wall x h - W_ladder x $$\frac{h}{2}$$cos$$\theta$$ - W_person x (2.7 x cos$$\theta$$) =0

The Attempt at a Solution

$$\theta$$ = cos^-1($$\frac{3.577}{7.3}$$) = 61

tan$$\theta$$61=$$\frac{h}{3.577}$$=6.45

3.65cos61=1.77

2.7 x cos61 = 1.31


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

N_wall x 6.45 -644 - 1061 = 0
N_wall x 6.45 = 1705
N_wall = $$\frac{1705}{6.45}$$
N_wall = 264 N

Its wrong. What am I doing wrong?

 

[auk411]

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)

 

[auk411]

by the way, L = height of ladder.

 

[DrMcDreamy]

I figured it out! Sorry for my messy work! :tongue:

 

[rwisz]

Your calculations for the normal force exerted by the floor on the ladder are correct. However, your calculation for the force exerted by the wall is incorrect. It seems like you have mixed up the variables in your equations. The correct equation for torque in this situation is Nwall x h - Wperson x (2.7 x cos\theta) - Wladder x \frac{h}{2}cos\theta = 0. Plugging in the values, we get Nwall x 3.577 - 810 N x 1.77 - 364 N x 1.31 = 0. Solving for Nwall gives us a value of 558 N, which is the correct answer for the force exerted by the wall. It is important to double check your equations and make sure you are using the correct values for each variable.