Question

please solve all

4) A 40.0 lb, 10.0 ladder leans against a smooth vertical wall as shown. It makes an angle of 60.0° with the horizontal floor. If the coefficient of static friction between the ladder and the floor is 0.400, how far up the ladder can a 180 lb fireman climb before sliding begins? 15 pts.) 5) A 150 kg mass is held in place on a rough incline plane (4,-0.350) by a horizontal force F as shown. Determine the maximum and minimun F that can be applied such that the mass does not slide up or down the incline. (You do not have to worry about the mass tipping over. Assume it is a point particle.) (15 pts.) F 6) Find the centroid (7.7) of the shaded region. (15 pts.) y=k 2 yox X 7) Find the centroid (%.9.2)of the paraboloid. 15 pts) Z Z= 3 by 9

Answer 1

4 Draw the free face Body diagram 40 lb NA 180 lb x let, t be the distance from from point B before sliding begins NS Vertical Equilibsicum, & Fy=0 NB-40-180 220 lb. Find the frictional force at B. MANB 1 BU Fea 29.4 x 220 FR = 88 16 Horizontal Equilê boricum, zfx=0 Fo-NA=0 F8 88 alebo let, distance AB is = 10 ft.

Take moment about B, EMB=0 to find distance a the 10 Cos 60 [Cow [MAX 10 Sin 68] - [HQ* (10 - [180] [2006] 8.66 NA 100.-90.x = 0 8-6€ X88-100 - 90x=0 x = 8.356 ff from point B the distance from point B firenas climb before sliding beging is 80356ff

F and the weight, W= ng 150 x 9.81 - 14T15N cose-la sliding up of mass 147165 SA Find the prictional farce, FA= Ms NA uthor =Qe 35 NA FA F Be 3& NA Resolve the forces along vertical Equilibrium, EF g = 0 sino =0 -147165+ NAC0838) - Fa -147105 + NA COS 38 -@35NA) Sin38 =0 NA = = 2129.44 N 2129.44 w

Resolve along horizontal directan 2Fx=0 + F NASIA 3o. - FA cos30=0 +FA = 0.5NA -0.303 NA = 0 11 F = 0.803 NA F= 0.803X 2129.44 F =11709.94 N marinum horizontal force u 1709.94N case 2: sliding down of mass 147165 FA Find the frictional farce, FA= Ms NA F 38 +0.35 NA Resolve the forces along vertical direction, Fy=0

-14715 + NA COS 38 + FA sin 30 =0 - 149 15+ NA (8838 +6:35 NA Sin 30=0 NA = 1413-51 = 1413.5 IN Horizontal Resove along weutical direction, IF, =0 F-NA SI 3 8 + Fa + FA C0838 = 0 cosso = 0 F-005 Na t 0.35 NA F = 001969 NA 0.1969 * 1413.51 3278.31 N minimum horizontal farbe is 278.310 6 y x "Centroid x distance. ca DE Let, fake small Element 2 y = 1 x dy and di t dx= x - x2 1 = 29 y

Now, Elementaal Area dA = drody = (29-4²) dy Find the centroid of the Element gła : 29+y [yty 2 2 Find the centroid of the total Areas Tecat fata lg ty (29-57 s da 2 29-g² jag y] dg ورم در 2 2 V { [247- 93 +y? (24-y? I dg { y3] To 16 m) = [1.6m) 93 95 2x 5 3 201333 2 [2.41 1.333

the Find the centroid of Ica- Find Elemental Area dA = dredy = (29-6?). Find the centroid y Element ☺ = y. Find the centroid 9 Area you sağda Cๆ J dd ly. (29-9") dy ' (@9-97) de $(2 g ² g 3 ay (29-42) dg 2 s 2. y3 3 1.333 4. IŚ 33 12. q کہا 3 1 you -1 M on 1 m

# Find the centroid ý distace given Equation, 2 ay Z= 3 ry 7 / 3 = Jy > z=ay Elemental volume, dv = (1 z") dy qy) dy Find the centroid ý ci distan y = 55.dv fy. ( 11sy hey Tin laylay Jar 9 dy 9 T Oo V yay He love No, whe Qt पु = 6 m Due to symmetry {=0, t=0 hence, the centroid, (X. 9, 3) = (0,6M, 10, 6, o

I have provided conceptually correct answer.

If you have any doubts please ask me.

Please provide positive rating and thank you.