Load on ladder leaning against a wall

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A physics problem involving a ladder leaning against a wall at a 30-degree angle is discussed, where an 800N load is placed three-quarters up the ladder. The user initially struggles to find the reaction force (R) and its angle (theta) due to insufficient equations. A response clarifies that three equations are needed for equilibrium: net forces in both x and y directions, and net torque. By taking moments about the ladder's base and denoting the ladder's length as L, the user successfully solves the problem. The discussion emphasizes the importance of considering moments and equilibrium in solving static problems.
engcon
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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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