An 88 kg person stands on a uniform ladder 4.0 m long, that weighs 80 N, as show

Question

An 88 kg person stands on a uniform ladder 4.0 m long, that weighs 80 N, as shown in the figure . The floor is rough; hence, it exerts both a normal force, f_1, and a frictional force, f_2, on the ladder. The wall, on the other hand, is frictionless; it exerts only a normal force, f_3.

a. Using the dimensions given in the figure, find the forces exerted on the ladder when the person is halfway up the ladder.

b. Find the forces exerted on the ladder when the person is three-fourths of the way up the ladder.

Explanation / Answer

a. The total forces are the ones mentioned, F_1, F_2, F_3 and using center of mass, gravity will always exert a force of 80N at the halfway point of the uniform ladder which is F_g, then the person standing halfway up the ladder will exert a force F_p for a total of 5 forces.

Directionally, F_1 is pointed straight up, F_2 is to the right, F_3 is to the left, F_g is down, and F_p is down.

Because the question is in static equilibrium, the forces must add up to 0 so separate the question into two parts.

X-comp: F_2 - F_3 = 0 ; F_2 = F_3

Y-comp: F_1 - F_g - F_p = 0 ; F_1 = F_g + F_p

F_2 = F_1 ; F_g = 80N and F_p = mg ; F_1 = F_g + mg ; F_1 = 80N + 88kg(9.8m/s^2)

F_2 = {80N + 88kg(9.8m/s^2)} = F_3

Torque: choose a torque point to eliminate the most possible torques, so choose the point of contact of the ladder to the floor. There is now three torques acting on the point, F_g, F_p, and F_3

= rFsin

of F_g is _g = 2m(80N)sin

of F_p is _p = 2m(88kg(9.8m/s^2))sin

of F_3 is _3 = 4m( {80N + 88kg(9.8m/s^2)})sin ; = 270 -

Again because of static equilibrium, = 0 and _g is in the negative Z direction as well as _p while _3 is in the positive Z direction, so _3 - _g - _p = 0 ; _3 = _g + _p

4m( {80N + 88kg(9.8m/s^2)})sin(270 - ) = 2m(80N)sin + 2m(88kg(9.8m/s^2))sin

b. The direction of the force caused by the person standing 3/4th of the way up of the ladder is still straight down, so the forces are exactly the same.

But the torque changes for _p, it becomes 3m(88kg(9.8m/s^2))sin because the position changed but the direction of the force is still the same.

torques is still _3 = _g + _p but new _p must be plugged in.

4m( {80N + 88kg(9.8m/s^2)})sin(270 - ) = 2m(80N)sin + 3m(88kg(9.8m/s^2))sin

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