1. 3.3 Coplanar Systems
A particle is subjected to coplanar forces in
the x-y plane
Resolve into i and j components for
equilibrium
∑Fx = 0
∑Fy = 0
Scalar equations of equilibrium
require that the algebraic sum
of the x and y components to
equal o zero

2. 3.3 Coplanar Systems
Scalar Notation
- Sense of direction = an algebraic sign
that corresponds to the arrowhead
direction of the component along each axis
- For unknown magnitude, assume
arrowhead sense of the force
- Since magnitude of the force is always
positive, if the scalar is negative, the force
is acting in the opposite direction

3. 3.3 Coplanar Systems
Example
Consider the free-body diagram of the
particle subjected to two forces
Assume unknown force F acts to the right for
equilibrium
∑Fx = 0 ; + F + 10N = 0
F = -10N
Force F acts towards the left for equilibrium

4. 3.3 Coplanar Systems
The chain exerts three
forces on the ring at A.
The ring will not move, or
will move with constant
velocity, provided the
summation of the forces
along the y axis is zero
With any force known, the
magnitude of other two
forces are found by
equations of equilibrium

5. 3.3 Coplanar Systems
Procedure for Analysis
1. Free-Body Diagram
- Establish the x, y axes in any suitable
orientation
- Label all the unknown and known
forces magnitudes and directions
- Sense of the unknown force can be
assumed

6. 3.3 Coplanar Systems
Procedure for Analysis
2) Equations of Equilibrium
- Given two unknown with a spring, apply
F = ks
to find spring force using deformation of
spring
- If the solution yields a negative result,
the sense of force is the reserve of that
shown in the free-body diagram

7. 3.3 Coplanar Systems
Procedure for Analysis
2) Equations of Equilibrium
- Apply the equations of equilibrium
∑Fx = 0 ∑Fy = 0
- Components are positive if they are
directed along the positive negative axis
and negative, if directed along the
negative axis

8. 3.3 Coplanar Systems
Example 3.2
Determine the tension in
cables AB and AD for
equilibrium of the 250kg
engine.

9. 3.3 Coplanar Systems
Solution
FBD at Point A
- Initially, two forces acting, forces
of cables AB and AD
- Engine Weight
= (250kg)(9.81m/s2)
= 2.452kN supported by cable CA
- Finally, three forces acting, forces
TB and TD and engine weight
on cable CA

10. 3.3 Coplanar Systems
Solution
+→ ∑Fx = 0; TBcos30° - TD = 0
+↑ ∑Fy = 0; TBsin30° - 2.452kN = 0
Solving,
TB = 4.90kN
TD = 4.25kN
*Note: Neglect the weights of the cables since they
are small compared to the weight of the engine

11. 3.3 Coplanar Systems
Example 3.3
If the sack at A has a weight
of 20N (≈ 2kg), determine
the weight of the sack at B
and the force in each cord
needed to hold the system in
the equilibrium position
shown.

12. 3.3 Coplanar Systems
Solution
FBD at Point E
- Three forces acting,
forces of cables EG
and EC and the
weight of the sack on
cable EA

13. 3.3 Coplanar Systems
Solution
+→ ∑Fx = 0; TEGsin30° - TECcos45° = 0
+↑ ∑Fy = 0; TEGcos30° - TECsin45° - 20N = 0
Solving,
TEC = 38.6kN
TEG = 54.6kN
*Note: use equilibrium at the ring to determine
tension in CD and weight of B with TEC known

14. 3.3 Coplanar Systems
Solution
FBD at Point C
- Three forces acting, forces by cable
CD
and EC (known) and
weight of sack B on
cable CB

15. 3.3 Coplanar Systems
Solution
+→ ∑Fx = 0; 38.6cos30° - (4/5)TCD = 0
+↑ ∑Fy = 0; (3/5)TCD – 38.6sin45°N – WB = 0
Solving,
TCD = 34.1kN
WB = 47.8kN
*Note: components of TCD are proportional to the slope
of the cord by the 3-4-5 triangle

16. 3.3 Coplanar Systems
Example 3.4
Determine the required length of the
cord AC
so that the 8kg lamp is suspended. The
undeformed length of the
spring AB is l’AB = 0.4m,
and the spring has a
stiffness of kAB = 300N/m.

17. 3.3 Coplanar Systems
Solution
FBD at Point A
- Three forces acting, force by cable AC,
force in spring AB and
weight of the lamp
- If force on cable AB is known,
stretch of the spring is found
by F = ks

18. 3.3 Coplanar Systems
Solution
+→ ∑Fx = 0; TAB – TACcos30° = 0
+↑ ∑Fy = 0; TABsin30° – 78.5N = 0
Solving,
TAC = 157.0kN
TAB = 136.0kN

19. 3.3 Coplanar Systems
Solution
TAB = kABsAB; 136.0N = 300N/m(sAB)
sAB = 0.453N
For stretched length,
lAB = l’AB+ sAB
lAB = 0.4m + 0.453m
= 0.853m
For horizontal distance BC,
2m = lACcos30° + 0.853m
lAC = 1.32m