Question

How do I solve this force/friction question?

Answer 1

Angle made by the ladder with horizontal
`theta = sin^-1(8.0/16) = sin^-1 0.5 = 30^@`

(a) There is no vertical force at the wall as wall is given to be friction less. Let `F_w` be the normal force at the wall.
As the system is in equilibrium sum the moments of all forces about the base must be `=0`. Remembering that weight acts at the centre of mass of the uniform ladder we get
`Sigma M = 0 = F_w xx8 - 210 xx16/2 xx cos30 - 40xx 2.0xx sin30`
`=>0 = 8F_w - 1454.9-40`
`=>F_w = 186.9 N`

Then `F_x = 186.9 - 40 = 146.9 N`

(b) `F_y = 210 N` (Simply `=` the weight of the ladder)

(c) Under the action of revised applied force
`Sigma M = 0 = F_w xx8 - 210 xx16/2 xx cos30 - 140xx 2.0xx sin30`
`=>0 = 8F_w - 1454.9-140`
`=>F_w = 199.4 N`

Then `F_x = 199.4 - 140 = 59.4 N`

(d) Same as in (b) above.

.-.-.-.-.-.-.-.-.-.-.-.-.-
This needs to be reviewed.

(e) Force of friction between the floor and ladder depends only on the normal force, which is `=210 N` for all cases.
`:.` Force of friction `F_f = muF_y = 0.40xx 210 = 84 N`

The ladder starts to move when `F_x= 84 N`.
Working backwards

`F = F_w-F_x = F_w-84` .....(1)

`Sigma M = 0 = F_w xx8 - 210 xx16/2 xx cos30 - Fxx 2.0xx sin30`

Using (1)
`Sigma M = 0 = F_w xx8 - 210 xx16/2 xx cos30 - ( F_w-84)xx 2.0xx sin30`
`=>0 = 8F_w - 1454.9 - (F_w-84)`
`=>7F_w = 1370.9`
`=>F_w = 195.8 N`
and finally

`F = 195.8 - 84 =111.8 N`

Answer 2

See below.

Explanation:

Calling

`theta_0 = arcsin(h/L)`

`A = (-L cos theta_0,0)`
`B = (0,L cos theta_0)`
`vec P = (0,-p)`
`vec F = (f,0)`
`vec R_A = (H_A,V_A)`
`vec R_B = (-H_B,0)`

we have for resultant

`vec F+vec P + vec R_A + vec R_B = vec 0`

and moments around `B`

`vec R_B xx (B-A) + vec P xx 1/2(B-A) + vec F xx (B-A) d/L = vec 0`

This way we obtain the equations

`{(f + H_A - H_B = 0),(V_A-p = 0),(L p Cos theta_0 = 2 (H_B L - d f) Sin theta_0):}`

and solving

`{(H_A = f (d/L-1) + 1/2 p Cot theta_0 ),(V_A=p),(H_B = (d f)/L + 1/2 p Cot theta_0):}`

putting values

a-b) `H_A = 146.865,V_A=210 = 176.865, H_B = 186.865`

c-d) `H_A = 41.865, V_A = 210, H_B = 201.865`

e) Supposing `mu = 0.40` and introducing the corresponding equation

`H_A = mu V_A`

and solving for `H_A,H_B,V_B, f`

`{(),(H_A=mu p),(V_A=p),(H_B=(2 d mu p - L p Cot theta_0)/(2 d - 2 L)),(f =(L p ( Cottheta_0-2 mu))/(2 (L-d))):}`

and putting values

`H_A=84, V_A = 210, H_B = 195.846,f->111.846`