The Student Room Group
Reply 1
Ive been trying to figure this one out, but I havent done it for about a year.

You can split the system into components, from a free body force diagram of it. You can then find the x component(vertical force) and y component(horizontal force) by using the information given.

Im not sure how moments can be used to find the forces involved though...
Saunders
6) A ladder of weight 250N and length 6.00m rests against a smooth vertical wall. Its top end is 5.00m above horizontal ground. A bucket of weight 100N hangs 0.500m from the top of the ladder and a person of weight 550N is 1.00m from the bottom of the ladder. Find the forces at the wall and at the ground.

Let x = Angle of elevation of the ladder from the horizontal:
sinx = o/a = 5/6
---> x = sin^-1(5/6) = 56.4 degrees (3.S.F)

There are 3 forces to consider in addition to the weights of the ladder, bucket and person.

1.) Normal reaction from the wall, which acts parallel to the ground. i.e.) Horizontally.
2.) Normal reaction from the ground, which acts perpendicular to the ground. i.e.) Vertically upwards.
3.) Friction acting opposite to the tendancy of motion, parallel to the ground. i.e.) Horizontally

Ladder is at rest, therefore Resultant (Vertical Component) = 0. Resolving Vertically:
---> Normal Reaction (From Ground) = 550 + 250 + 100
---> Normal Reaction (From Ground) = 900 N

There are 2 horizontal components; the friction at the ground and the normal reaction from the wall.
From geometry: The Normal Reaction from the wall is at an angle x from the ladder (Alternate Angles).

Ladder is at rest therefore Resultant (Horizontal Component) = 0. Resolving Horizontally:
---> Friction = Normal Reaction (From Wall)

Let A = The Point at which the ladder makes contact with the ground.

* We assume the ladder is uniform and thus the weight of the ladder is said to act at its centre of mass at the midpoint of the ladder. i.e.) 3 m from both ends.

Taking Moments about A:

---> (550*1)cosx + (250*3)cosx + (100*5.5)cosx = 6*Normal Reaction (From Wall)*sinx
---> Normal Reaction (From Wall) = (550cos56.4 + 750cos56.4 + 550cos56.4)/[6(sin56.4)]
---> Normal Reaction (From Wall) = 204.9 N (4.S.F)

Hence: ---> Friction (At Ground) = 204.9 N (4.S.F)

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