ME 301 – MECE 301 THEORY OFD MACHINES I
SOLVED PROBLEM SET 3
PROBLEM 1
A one-d.o.f. mechanism is shown which involves two Revolute joints and one Cylinder in
Slot joint. The system is at static equilibrium under the effects of force R and torque T. Line
of action of the force R has an angle 𝛼 withe the horizontal as shown. Neglect gravitational
forces. The fixed link dimensions are 𝐴𝑜 𝐵𝑜 = 𝑟1, 𝐵𝑜 𝐶 = 𝑟3, 𝐵𝑜 𝐴 = 𝑏3 , ∡𝐴𝐵𝑜 𝐵 = 𝛾)
a) Draw the free body diagrams of the links using the link figures provided
b) Write the necessary equations for each rigid body from which T and the reaction
forces can be solved in terms of the joint variables (𝜃12 , 𝑠23 and 𝜃13 ), the fixed link
dimensions and the given force R.
Link 2
Link 4
Solution:
Link 3:
∑ 𝑀𝐵𝑜 = 0
∑ 𝐹𝑥 = 0
∑ 𝐹𝑦 = 0
⇒
𝑅𝑟3 sin(𝛼 + 180𝑜 − 𝜃13 ) + 𝐹23 𝑏3 sin(𝜃12 + 270𝑜 − 𝜃13 − 𝛾) = 0
⇒
−𝑅𝑟3 sin(𝛼 − 𝜃13 ) + 𝐹23 𝑏3 cos(𝜃12 − 𝜃13 − 𝛾) = 0
⇒
𝐹23 =
𝑏3 cos(𝜃12 −𝜃13 −𝛾)
𝑅𝑟3 sin(𝛼−𝜃13 )
⇒
𝑥
𝐹13
+ 𝑅 cos(𝛼 + 180𝑜 ) + 𝐹23 cos(𝜃12 + 270𝑜 ) = 0
⇒
𝑥
𝐹13
= 𝑅 cos 𝛼 − 𝐹23 sin𝜃12
⇒
𝐹13 + 𝑅 sin(𝛼 + 180𝑜 ) + 𝐹23 sin(𝜃12 + 270𝑜 ) = 0
⇒
𝐹13 = 𝑅 sin 𝛼 + 𝐹23 cos𝜃12
𝑦
𝑦
𝐹23 = 𝐹32 = 𝐹12
Link 2:
∑ 𝑀𝐴𝑜 = 0
⇒
𝑇 + 𝐹32 𝑠23 = 0
⇒
𝑇 = −𝐹32 𝑠23
a)
b)
PROBLEM 3:
In the six-link mechanism shown the joint variables are 𝜃12 , 𝜃13 , 𝜃14 , 𝑠54 and 𝑠16 . Load 𝑃 is
known. Gravitational, frictional and inertial effects are negligible. Link dimensions: 𝐴𝑜 𝐴 =
𝑎2 , 𝐴𝐵 = 𝑎3 , 𝐵𝑜 𝐵 = 𝑎4 , 𝐶𝐷 = 𝑏6 , 𝐷𝐸 = 𝑐6 , ∠𝐻𝐵𝑜 𝐵 = 𝛽4 and other dimensions are shown
in the figure.