- #1

moenste

- 711

- 12

## Homework Statement

A uniform ladder which is 10 m long and weights 300 N leans with its upper end against a smooth vertical wall and its lower end on rough horizontal ground. The bottom of the ladder is 6 m from the base of the wall. A man weighting 700 N stands on the ladder at a point 6 m above the ground. Calculate the magnitudes and directions of the forces exerted on the ladder by (a) the wall, (b) the ground.

Answers: (a) 506 N at 90 degrees to the wall, (b) 1121 N at 63 degrees to the ground.

**2. The attempt at a solution**

At first I did a graph of the situation:

R = normal reaction to the wall, S = to the ground, F = friction (rough ground).

Find 8 by Pythagoras. S is = 700 N + 300 N = 1000 N.

And after this point I am a bit stuck. From a different book example I get:

(F * 8) + (300 * 3) + (700 * 1.5) = (S * 6)

8 F = 4050

F = 506.25 N

The answer looks right, but:

As I understand, the body is in equilibrium so in that case: normal reaction S (anti-clockwise) should be multipled by the perpendicular line which is 6 m -> (S * 6), then all of other forces which are facing clockwise direction should be used: we have F and we sort of move it to the left so the perpendicular line is 8 m (F * 8), two other clockwise forces are 700 and 300. 300 * 3 and 700 * 1.5 (the vertical side of the triangle is 2 m, the hypotenuse is 2.5 m and by Pythagoras the horizontal line is 1.5). So we get F = 506 N. But how to find the angle? Because F is a force directed towards the wall that is the reason why it is equal to 90? And why don't we take into account the R force?

And how to start with (b)?

Any help please?