Load on ladder leaning against a wall

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The discussion revolves around solving a physics problem involving a ladder leaning against a wall, specifically analyzing the forces and moments acting on the system. The ladder is positioned at a 30-degree angle with an 800N load placed three-quarters of the way up. The key equations for equilibrium include Fnet = 0 in both x and y directions, and Mnet = 0 for torque. By denoting the ladder's length as L, participants successfully derived the necessary equations to find the reaction force (R) and its angle (theta).

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Hello, I got this problem from a book

"A light ladder leans against a perfetly smooth vertical wall at an angle of 30 degrees to the horizontal. A load of 800N is placed 3-quarters of the way up the ladder. If the ladder rests on a rough horizontal surface which prevents slipping, find the magnitude and direction of the reaction between the ladder and the ground"

Since the system is in equilibrium, I resolved vertically and got

R sin(theta) = 800

where R is the reaction force, and theta is the angle of the reaction force with the horizontal.

The problem is I can't get another equation so that I can find R and theta.

Any help is appreciated, thanks!
 
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Okay, first of all, my definition of "reaction" is the normal force produced by any surface, which I think does not coincide with your definition. In anycase, This problem involves three equations which are easily solved.

1. Fnet = 0 (in the x direction)
2. Fnet = 0 (in the y direction)
3. Mnet = 0 (torque, no rotation)

The frictionless wall will only produce a force in x. The floor will only produce a force in y. The friction at ladder's base will produce a force in -x. The 800 N load will produce a force in the -y.

Now for moments. Take the moments about the ladder's base (you can choose any point, but this point is particularly easy. Then the two forces that will create a torque around this point will be the 800N load (counter clockwise) and the reaction force of the wall (clockwise). They must sum up to zero.

Go get em =)
 
mezarashi said:
Now for moments. Take the moments about the ladder's base (you can choose any point, but this point is particularly easy. Then the two forces that will create a torque around this point will be the 800N load (counter clockwise) and the reaction force of the wall (clockwise). They must sum up to zero.

Go get em =)

No mention of the ladder length is made, so I cannot use moments
 
engcon said:
No mention of the ladder length is made, so I cannot use moments

Are you sure? Try denoting the length of the ladder as L. Leave it as L (you don't need to substitute a number), and see what happens.
 
Solved

Did that, and solved it

Thanks
 

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