Department of Mechanical Engineering
Assignment: 04
COURSE CODE
: ME 249
COURSE NAME
: Engineering Mechanics-II
STUDENT NAME
: Rezvi Sannamat
STUDENT ID
: BME2001020077
SECTION
: ME-0121-Denali
DEPARTMENT
: Mechanical Engineering
Q The coefficient of friction between blocks
A and C and the horizontal surfaces is
m_k = 0.2X. Knowing that m_A = 5.D kg,
m_B = 1X kg, and m_C = 10 kg,
determine (a) the tension in the cord, (b)
the acceleration of each block.
[D is the last digit of your birth day and X is
the last digit of your ID]
Solution:
From the figure block (A) we get :
∑𝐹# : 𝑅& = 𝑚𝐴 𝑔
= 5.9 𝑘𝑔 × 9.8 𝑚/𝑠2 W=𝑚& 𝑔
= 28.42 𝑁
𝐹& = 𝜇𝐾 𝑅&
= 0.27 × 28.42 𝑁
= 7.67 N
𝐹& = 𝜇= 𝑅&
D is the last digit of your birth
day
𝑇
𝑚& 𝑎 &
𝑅&
@
→ ∑𝐹𝑥 = 𝑚& 𝑎&
T-28.42 𝑁 = 5.9 𝑘𝑔 × 𝑎𝐴
FGHI.JH K
𝑎& =
L.M NO
D=9
X is the last digit of your ID
X= 7
𝜇= = 0.2𝑋
𝜇= = 0.27
𝑚& = 5.9 𝑘𝑔
𝑚C = 17 𝑘𝑔
𝑚E = 10 𝑘𝑔
𝑔 = 9.8 𝑚/𝑠 H
𝑎& = 0.16 T − 4.81 ----------------(1)
From the figure block (B) we get
+↓ ∑𝐹𝑦 = 𝑚C 𝑎C
𝑚C 𝑔 − 2𝑇 = 𝑚𝐵 𝑎𝐵
17 kg × 9.8 𝑚/𝑠 H − 2𝑇 = 17 𝑘𝑔 × 𝑎C
𝑎C =
VW XY ×M.I Z/[ \ GHF
VW XY
.
𝑎C = 9.8 − 0.11𝑇 ----------------(2)
From the figure block (C) we get
∑𝐹# : 𝑅C = 𝑚𝐵 𝑔
= 10 𝑘𝑔 × 9.8 𝑚/𝑠2
= 98 𝑁
2T
𝑚C 𝑎C
𝐹E = 𝜇𝐾 𝑅E
= 0.27 × 98 𝑁
= 26.46𝑁
W=𝑚E 𝑔
.
𝑚E 𝑎E
T
@
← ∑𝐹𝑥 = 𝑚E 𝑎E
𝑇 − 26.46𝑁 = 10 𝑘𝑔 × 𝑎E
FG26.46𝑁
𝑎E =
𝑅E
V^ XY
𝑎E = 0.1𝑇 − 2.64 ----------------(3)
Kinetic acceleration ,
V
𝑎C = (𝑎& + 𝑎E )
H
a) The tension in the cord,
We take the value kinetic accelerations equation (1), (2), (3)
𝑎C =
V
H
(𝑎& + 𝑎E )
V
9.8 − 0.11𝑇 = (0.16 T − 4.81 + 0.1𝑇 − 2.64)
H
19.6 − 0.22𝑇 = 0.26 T − 7.45
HW.^L
𝑇=
^.JI
𝑻 = 𝟓𝟔. 𝟑𝟓 𝑵 (Ans )
b) Find out acceleration each blocks,
We get , 𝑇 = 56.35 𝑁
𝑎& = 0.16 T − 4.81
= 0.16 × 56.35 𝑁 − 4.81
= 𝟒. 𝟐𝟎 𝒎/𝒔𝟐 (Ans )
𝑎C = 9.8 − 0.11𝑇
= 9.8 − 0.11 × 56.35 𝑁
= 𝟑. 𝟔𝟎 𝒎/𝒔𝟐 (Ans )
𝑎E = 0.1𝑇 − 2.64
= 0.1 × 56.35 𝑁 − 2.64
= 𝟑. 𝟏𝟕 𝒎/𝒔𝟐 (Ans )
𝐹E = 𝜇= 𝑅E