mechanics tutorial

1. 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
= ΣF = 25Cos(60 ) − 40 = −27.5 (←)
= ΣF = 25Sin(60 ) − 10 = 11.6506 (↑)
= = 80 + (10)(12) − 40(8) = −120 . cm (clockwise)
Resultant Force
→
= ̂ + ̂ = (−27.5 ) ̂ + (11.6506 ) ̂
= |
→
| = 29.866 N
b) Line of action of the single resultant force
i) Intersection with line AB
Hence (11.6506) = 120 ⇒ = 10.3cm

2. ii) Intersection with line BC
(27.5) = 120 ⇒ = 4.3636 cm
Single Resultant Force

3. 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  
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4. 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
(3) − (ℎ − 4)(10)(6)
10
2
= 0 => ℎ = 4.5
R
mg
Fp

5. 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?
( ) = ρgz
= ρgz 2R(z) ds
Note that
dz
ds
= sin
( )
=
1
tan
=> ( ) = ⁄
Thus, total force in vertical direction is
= ∫ dR cos
= ∫ ρgz ds cos
2πz
tan
= ∫ 2πρgz
dz
sin
cos
1
tan
dz
=
2πρg
(tan )
dz
=
2πρg
2
∗
2
3
=
4πρg
3
= 41.8 kN
Which is equal to the minimum weight
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