Tutorial-1 Solution Problem 1 The three forces and a couple shown are applied to an angle bracket. (a) Reduce the system to a force and a couple at B. (b) Locate the points where the line of action of the single resultant force intersects line AB and line BC.

a) Reduce the system to a force and a couple at B

푅 = ΣF = 25Cos(60 ) − 40 = −27.5푁(←)

푅 = ΣF = 25Sin(60 ) − 10 = 11.6506푁(↑)

푀 = 훴푀 = 80 + (10)(12) − 40(8) = −120푁. cm(clockwise)

ResultantForce푅→

= 푅 횤̂ + 푅 횥̂ = (−27.5푁)횤̂ + (11.6506푁)횥̂

푅 = |푅→

| = 29.866N

b) Line of action of the single resultant force i) Intersection with line AB

Hence(11.6506)푥 = 120 ⇒ 푥 = 10.3cm

ii) Intersection with line BC

(27.5)푦 = 120 ⇒ 푦 = 4.3636cm

Single Resultant Force

Problem 2

Three children are standing on a 15 15-ft raft. The weights of the children at points A, B, and C are 85 lb, 60 lb, and 90 lb, respectively. If a fourth child of weight 95 lb climbs onto the raft, determine where she should stand if the other children remain in the positions shown and the line of action of the resultant of the four weights is to pass through the center of the raft.

Have : A B C D F F F F F R

85 lb 60 lb 90 lb 95 lb j j j j R

330 lb R j

Have :x A A B B C C D D HM F z F z F z F z R z

85 lb 9 ft 60 lb 1.5 ft 90 lb 14.25 ft 95 lb 330 lb 7.5 ftDz

3.5523 ftDz or 3.55 ftDz

Have :z A A B B C C D D HM F x F x F x F x R x

85 lb 3 ft 60 lb 4.5 ft 90 lb 14.25 ft 95 lb 330 lb 7.5 ftDx

7.0263 ftDx or 7.03 ftDx

Problem 3. Gate AB shown in the figure below is 6m wide and weighs 50,000 kg when submerged. It is hinged at B and rests against a smooth wall at A. Determine the water level h which will just cause the gate to open (see Fig. 2).

R is resultant of FP and mg R = -mg ĵ + ρg(h-4)(10)(6)[ cos Ɵ î + sin Ɵ ĵ ] cos Ɵ = 0.8 , sin Ɵ = 0.6 R = ρg(h-4)(48)î + [ρg(h-4)(16) – mg]ĵ Impending opening of gate when line of action of R coincides with AB: => ( )( )

( )( )= − =>ρg(h− 4)[36 ∗ 6 − 48 ∗ 8] = 6mg => h = 4.5m

Check by equilibrium approach: Put ∑MB = 0 with reaction at A = 0

푚푔(3)− 휌푔(ℎ − 4)(10)(6)102 = 0 => ℎ = 4.5푚

R

mg

Fp

Problem 4. A hollow steel cone with internal dimensions as shown has a pinhole at the top. The cone is filled with water. What is the minimum weight of the cone which will prevent the water from up-lifting the cone and flowing out?

푝(푧) = ρgz 푑푅 = ρgz2R(z)ds

Note that dzds = sin훼 푅(푧)푧 =

1tan훼 => 푅(푧) = 푧 푡푎푛훼⁄

Thus, total force in vertical direction is 퐹 = ∫ dRcos훼

= ∫ ρgzdscos훼2πztan훼

= ∫ 2πρgzdz

sin훼 cos훼1

tan훼 dz

=2πρg

(tan훼) 푧 dz

=2πρg

2 ∗23 =

4πρg3 = 41.8kN

Which is equal to the minimum weight