Tutorial - 12 (solutions)
Question 1 The 7.5-kg rod BC, see Fig. (1), connects a disk centered at A to crank CD. Knowing that the disk is made to rotate at the constant speed of 180 rpm, determine
for the position shown the vertical components of the forces exerted on rod BC by
pins at B and C. (15 Marks)
Figure 1: Figure for question 1.
Solution 1 We first determine the acceleration of point B of disk. ω = 180 rpm = 18.85 rad/s.
1 mark for finding
aB
Since ω = constant, aB = (aB )n = rω 2 = 71.0645 m/s2 at an angle of 60○ with the
horizontal directed from B to A ↙.
2
○
1 mark for finding a Since rod BC is in translation, ā = āB = 71.0645 m/s at an angle of 60 with the
horizontal ↙.
Now the free body diagram and equivalent kinetic diagram for rod BC is shown
below
3 marks for
correctly making
FBD and KD
Figure 2: Free body and equivalent kinetic diagram of BC.
Vertical components of forces at B and C.
1 mark for this step + Á ΣMB = Σ(MB )eff : Cy × 750 − W × 375 = −m ā sin60○ × 375
Now, W = 7.5 × 9.81 = 73.575 N and m ā = 532.9837 N.
1 mark for writing
this equation
∴ 750 × Cy − 73.575 × 375 = −532.9837 sin60○ × 375
2Cy = 73.575 − 461.5775
Cy = −194.0013 N (↓). 3 marks for finding Cy
1 mark for this step + ↑ ΣFy = Σ(Fy )eff : By − W + Cy = −m ā sin60○
Solving we get By = −194.0012 N (↓). 3 marks for finding By
1
1 mark for writing
this equation
Question 2 Each of the 3-kg bars AB and BC, see Fig. (1), is of length L = 0.5 m. A horizontal
force P of magnitude 20 N is applied to bar BC. Determine the angular acceleration
of each bar when b = L.(25 Marks)
Figure 1: Figure for question 2.
Solution 2 Kinematics: Assume αAB Á and αBC Á. Also, ωAB = ωBC = 0
4 marks for this step
i.e.,
(i) Making the diagram
(ii) Finding aAB, aB, and
aBC
Kinetics:
Bar BC:
3 marks for this step
¯ BC + (māBC ) L
+ Á ΣMB = Σ(MB )eff : P L = Iα
2
P =
1 mark for this step
1
1
m L αAB + m L αBC 1 mark for this step (1)
2
3
2
1 mark for this step
+ → ΣFx = Σ(Fx )eff : P − Bx = māBC
P − Bx = m (L αAB +
1
L αBC ) 1 mark for this step (2)
2
Bar AB:
3 marks for making
this FBD and KD
¯ AB + māAB L
+ Á ΣMA = Σ(MA )eff : Bx L = Iα
2
Bx =
1 mark for this step
1
m L αAB .
3
1 mark for this step
(3)
Adding (2) and (3):
P =
4
1
m L αAB + m L αBC . 1 mark for this step (4)
3
2
Subtract (1) and (4) we get
αBC = −5 αAB .
1 mark for this step
(5)
Substitute for αBC in (1):
7
P = − mLαAB .
6
1 mark for this step
6 P
.
7 mL
1 mark for this step
(6)
30 P
.
7 mL
1 mark for this step
(7)
αAB = −
Using Eq. (5)
αBC =
Using: L = 500 mm = 0.5 m, m = 3 kg, P = 20 N, we get
2
αAB = −11.4286 rad/s ⤸ . 2 marks for correctly finding
out this
2
αBC = 57.1429 rad/s Á .
3
2 marks for correctly finding
out this
Question 3 A uniform slender bar is released from rest in the horizontal position shown. Determine the value of x for which the angular acceleration is a maximum and determine
the corresponding value of angular acceleration α.(10 Marks)
Figure 1: Figure for question 3.
Solution 3 The free body diagram of the bar is shown below.
2 marks for making
FBD
Figure 2: Free body diagram of the bar.
Now, IO = IG + m x2 =
1
2
12 ml
2
l
+ mx2 = m ( 12
+ x2 )
2
l
+ Á ΣMO = IO α: m g x = m ( 12
+ x2 ) α
Solving for α, we get
α=
l2
( 12
1 mark for finding IO
1 mark for finding MO
gx
+ x2 )
1 mark for this step
∴ For maximum value of α we get
dα
= 0.
dx
1 mark for this step
This implies that
l
x= √ .
2 3
Substitution of this value of x in expression for α we get
√ g
α= 3 .
l
4
2 marks for this step
2 marks for this step