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Two boxes (see below) are linked rigidly and placed on an incline with an angle of 35Β°. The coefficients of kinetic friction are 0.12 and 0.28 between the inclined plane and Blocks 1 and 2, respectively. The masses of the blocks are 9 π‘˜π‘” and 6 π‘˜π‘”, respectively. Assuming that the mass of the link is negligible compared to the mass of the blocks, a. Determine and plot the acceleration, velocity, and position of the two block system as functions of time. b. What is the force (in vector form) exerted on Block 1 by Block 2?

Question-AnswerCategory: Engineering MechanicsTwo boxes (see below) are linked rigidly and placed on an incline with an angle of 35Β°. The coefficients of kinetic friction are 0.12 and 0.28 between the inclined plane and Blocks 1 and 2, respectively. The masses of the blocks are 9 π‘˜π‘” and 6 π‘˜π‘”, respectively. Assuming that the mass of the link is negligible compared to the mass of the blocks, a. Determine and plot the acceleration, velocity, and position of the two block system as functions of time. b. What is the force (in vector form) exerted on Block 1 by Block 2?
Rajesh kumar asked 7 months ago

Two boxes (see below) are linked rigidly and placed on an incline with an angle of 35Β°. The coefficients of kinetic friction are 0.12 and 0.28 between the inclined plane and Blocks 1 and 2, respectively. The masses of the blocks are 9 π‘˜π‘” and 6 π‘˜π‘”, respectively. Assuming that the mass of the link is negligible compared to the mass of the blocks, a. Determine and plot the acceleration, velocity, and position of the two block system as functions of time. b. What is the force (in vector form) exerted on Block 1 by Block 2?

1 Answers
R K answered 7 months ago

Β 
 

Consider
N₁
APAHT
fs2
let
-
=
mig
176-6
Ni= mig сого
Nβ‚‚
FSI = UKINI
Est
26 FST
-fst =
√3+
F52
FSβ‚‚
F52 =
Nβ‚‚
=
a
Mβ‚‚ g Bosq
systAccording to
to Newten Law - (F = ma)
(M₁ + mβ‚‚) a = (m₁ g Sma + mβ‚‚ 9 smo - Fs₁ - FS2)
9x9.81x5m35+6x9.81x Sm35-8.68-13-5
(0)
Velocity Us time I
ve
62.22 t
at
at
t=0
1.86.66
δΈͺ
δΈͺ
Sem
t
v
t = 2
+= 3
-124.94
62.22-
Position
= 1
+=o
t = 1
279.99
+=2
t = 3- let force Exerted on Block I by Block Β· 2
be F
FOD of
Ans.
As
F
q
P
According to
m,g SmQ tf = fst)
m, q
9x 9.81 x Sm 35 +F

Your Answer

12 + 6 =