Answer:(a)360N,(b)171N,(c)2.702m
Explanation:
(a)Maximum Friction Force =[tex]\mu \left ( N\right )=0.4\times \left ( 740+160\right )[/tex]
=360 N
[tex]cos\theta =\frac{3}{5}[/tex]
[tex]sin\theta =\frac{4}{5}[/tex]
(b)Moment about Ground Point
[tex]740\times 1\times cos\theta +2.5\times 160\times cos\theta -N_15sin\theta [/tex]
[tex]N_1tan\theta =1140[/tex]
[tex]N_1=171 N[/tex]
[tex]N_1=f=171 N[/tex]
(c)
[tex]740\times x\times cos\theta +2.5\times 160\times cos\theta -N_15sin\theta [/tex]
Here maximum friction force can be 360 N
Therefore [tex]N_1=360 N[/tex]
Where x is the maximum distance moved by man along the ladder
[tex]360\times 5\times \frac{4}{3}=740x+160\times 2.5[/tex]
740x=2000
x=2.702m
The maximum friction force, actual friction and distance of the ladder are respectively
Formula for finding friction is given as
F = μN, or in this case, F = μ(N1 + N2)
F = 0.40(740 + 160)
F = 0.40(900)
F = 360 N
Look at the attached image, we can agree that from the triangle,
cosθ = [tex]\frac{3}{5}[/tex]
sinθ = [tex]\frac{4}{5}[/tex]
if so, then the moment about the ground point
740 * 1 * cosθ + 2.5 * 160 * cosθ - 5 * n * sinθ
740 * 0.6 + 2.5 * 160 * 0.6 - 5n * 0.8
444 + 240 - 4n
684 - 4n = 0
4n = 684
n = 684/4
n = 171 N
For the last part,
740 * x * cosθ + 2.5 * 160 * cosθ - 5 * n * sinθ
If the maximum friction force, n is 360, then
740x * 0.6 + 2.5 * 160 * 0.6 - 5 * 360 * 0.8
444x + 240 - 1440
444x - 1200
444x = 1200
x = 1200/444
x = 2.7m
to read more about kinematics of motion, see https://brainly.com/question/13671823
