Engineering Mechanics – Equilibrium of Force System | Fully Solved Problems (Step-by-Step Solutions)

Problem No. 2/1: The magnitude of force F is 500 N. Express F as a vector in terms of the unit
vector i and j. Identify both the scalar and vector component of F.

Solution:
Here,
Fx = -500 cos60 = -250N
Fy = 500 sin60 = 433N
Again,
F = Fxi + Fyj
or, 500 = (-250i + 433j) N
so, Fx = -250i and Fy = 433 j



Problem No. 2/2: When the load L is 7m from the pivot at C, the tension, T in the cable has a
magnitude of 9 kN. Express T as a vector using the unit vectors i and j.

Solution:
Here, tan = 6/10
So,  = tan-1 (0.6) = 310
Tx = 9 cos 31 = 7.7 kN
Ty = 9 sin 31 = 4.63 kN
So, T = Txi + Tyj
= ( 7.7 i + 4.63 j)

,Problem No. 2/10: Determine the scalar components Ra and Rb along the rectangular axes a and b.
Also determine the orthogonal projection Pa of R onto axis a.

Solution:




From Force Diagram, Applying Sine rule

Ra Rb R
= =
Sin 110 Sin 30 Sin 40 0
Ra R 800
 0
= 0
=
Sin 110 Sin 40 Sin 40
R  Sin 110 800  Sin 110
 Ra = =
Sin 40 Sin 40
 Ra =1169.5 N  1170 N
R  Sin30 0 800  Sin 110
Rb = =
Again, Sin 40 Sin 40
 Rb = 622.3 N  622 N

From the Figure
Pa = RCos30 = 800 Cos 30
 Pa = 693N

Problem No. 2/11: The 20-kN force is to be replaced by two forces F1, directed along the axes a-a,
and F2, which has a magnitude of 18 kN. Determine the magnitude of F1 and
angle , which F2 makes with the horizontal.




Solution: From Force diagram
F1 Cos 60 + 18 Cos 1 = 20 ........................ (1)
And, F1 Sin 60 – 18 Sin 1 = 0 .............................(2)

, or, F1 Sin 60 = 18 Sin 1
18 Sin  1
or, F1 = Putting value of F1 in (1)
Sin 60

18 Sin  1Cos 60
From (1), + 18 Cos  1 = 20
Sin 60
20
or, Sin 1 Cos 60 + Cos 1 Sin 60 = Sin 60
18
or, Sin (1 + 60) = 0.962
or, 1 + 60 = Sin-1(0.962) = 74.2
or, 1 = 74.2 – 60 = 14.20
Again, Sin (1800 – (2+60)) = Sin 74.2
or, 2 = 180 – 60 – 74.2 = 45.80
1 = 14.20 and 2 = 45.80
18 Sin 14.2
So, For 1 = 14.20, F1 = = 5.1 kN
Sin 60
18 Sin 45.8
So, For 2 = 45.80, F1 = = 14.9 kN
Sin 60

Problem No. 2/12: Repeat problem No. 2/11, except let F2 have a magnitude of 25-kN.

Given, F2 = 25 kN, F1 = ?

a. From Fig. a
F1 Cos 60 + 25 Cos  = 20 .................. (1)
And F1 Sin 60 – 25 Sin  = 0 ................... (2)
25 Sin 
F1 = ............................. (3)
Sin 60
From (1)
25 SinCos 60
+ 25 Cos = 20
Sin60
20
 SinCos 60 + CosSin60 = Sin60 0
25
 Sin( + 60) = 0.6928 = Sin 43.2
 Sin 180 − ( + 60) = Sin43.2
180 − ( + 60) = 43.2
 = 76.2 0
25 Sin 76.2
F1 = = 28 kN
Sin 60