#### Notes on Basics of Mechanics for GATE Mechanical Engineering-Free Body Diagrams

Related Free-Body Diagrams

When two or more bodies are in contact, separate free-body diagrams may be drawn for each body. The mutual forces and moments between the bodies are related according to Newton’s third law (action and reaction). The directions of unknown forces and moments may be arbitrarily assumed in one diagram, but these initial choices affect the directions of unknowns in all other related diagrams. The number of unknowns and of usable equilibrium equations both increase with the number of related free-body diagrams.

Schematic Example in Two Dimensions

Figure 1

Given: $F_{1},&space;F_{2},&space;F_{3},&space;M$

Unknowns: $P_{1},&space;P_{2},&space;P_{3}$ and forces and moments at joint A (rigid connection)

Equilibrium Equations

$\sum&space;F_{x}=-F_{1}+P_{3}=0$

$\sum&space;F_{y}=P_{1}+P_{2}-F_{2}-F_{3}=0$

$\sum&space;M_{0}=P_{1}c+P_{2}(c+d+e)+M-F_{2}a-F_{3}(a+b)=0$

Three unknowns $(P_{1},&space;P_{2},&space;P_{3})$ are in three equations.

Related Free-Body Diagrams

Figure 2

Dimensions a, b, c, d and of Figure 1 are also valid here.

New set of Equilibrium Equations

Left Part (OA):

$\sum&space;F_{x}=-F_{1}+A_{x}=0$

$\sum&space;F_{y}=P_{1}+A_{y}-F_{2}=0$

$\sum&space;M_{0}=P_{1}c+A_{y}(c+d)+M_{A}-F_{2}a=0$

Right Part (AB):

$\sum&space;F_{x}=-A_{x}+P_{3}=0$

$\sum&space;F_{y}=P_{2}-A_{y}-F_{3}=0$

$\sum&space;M_{0}=-M_{A}+P_{2}e+M-F_{3}f=0$

Six unknowns $(P_{1},&space;P_{2},&space;P_{3},&space;A_{x},&space;A_{y},&space;M_{A})$ are in six equations.

Example:

The arm of a factory robot is modeled as three bars, as shown in the figure below with coordinates A: (0.6, -0.3, 0.4) m; B: (1, -0.2, 0) m: and C: (0.9, 0.1, -0.25) m. The weight of the arm is represented by $W_{A}=-60Nj$ at A, $W_{B}=-40Nj$ at B. A moment $M_{C}=(100i-20j+50k)N.m$ is applied to the arm at C. Determine the force and moment reactions at O, assuming that all joints are temporarily fixed.

Solution:

The FBD is drawn in the figure (b) above, showing the unknown force and moment reactions at O.

$\sum&space;F=0$

$F_{o}+W_{A}+W_{B}=0$

$F_{o}-60Nj-40Nj=0$

$F_{o}=100Nj$

$\sum&space;M_{o}=0$

$M_{o}+M_{C}+(r_{OA}\times&space;W_{A})+(r_{OB}\times&space;W_{B})=0$

$M_{o}+100N.m&space;\bold&space;{i}-20N.m&space;\bold&space;{j}+50N.m&space;\bold&space;{k}-36N.m\bold&space;{k}+24N.m\bold&space;{i}-40N.m\bold&space;{k}=0$

$M_{O}=(-124\bold{i}+20\bold{j}+26\bold{k})&space;N.m$

An example for you to try out:

A load of 7kN may be placed anywhere within A and B in a trailer of negligible weight. Determine the reactions at the wheels at D, E and F, and the force on the hitch H that is mounted on the car, for the extreme positions A and B of the load. The mass of the car is 1500 kg and its weight is acting at C.

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