The question concerns a ladder that is 15.0 m long weighing 500N isresting again

Question

The question concerns a ladder that is 15.0 m long weighing 500N isresting against a frictionless wall and is making an angle 60.0degrees above the ground with a firefighter 4.00 m from the bottomand weighing 800N.

When figuring out the horizontal force exerted on the base of theladder, the equation is:
= 0 = (800N)sin30.0(4m) + (500N)sin30.0(7.5m) - (normalforce of the wall)cos30.0(15.0m)

I don't understand why we use sin30.0 degrees instead of sin 60.0degrees for the force exerted by the fireman and the ladder, andnot cos60.0 instead of sin60.0 for the normal force exerted by thewall.

Explanation / Answer

The ladder makes a 60 degree angle with the ground. That means theforce of the man and the ladder (pointing directly down, in thenegative y direction) both make a 30 angle with the ladder. So thatis the angle you need to figure out the horizontal force on thebase of the ladder. sin = o/h so you are solving for the forceopposite the angle, which is the ground. Same idea for the force onthe wall.

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