ENS 161 STATICS of RIGID BODIES
Chapter 1: Statics of Particles
2nd Semester A.Y. 2024 - 2025
Forces as Vectors and Resultants
● General
Forces are drawn as directed arrows. The length of the arrow represents the magnitude of the force and
the arrowhead determines its direction. Forces on rigid bodies further have a line of action.
Sample Problem: Force Components in Coplanar (2D) Force Systems
1. Given a 30° - inclined plane with
a block weighing 100N, as
shown. Assuming the block is in
equilibrium
(no
downward
motion), find the components of
the gravity force on the block
normal
and
parallel
to
the
Figure 1 Inclined Plane
inclined plane.
SOLUTION 1: By geometry, the angle between F and Fn is 30°, and that between F and Fp is 60°.
Using the definition of the trigonometric function sine and cosine, gives,
Fn = 100 cos 30° or 86.603 N ; Fp = 100 cos 60° or 50.000 N
Sample Problem: Magnitude and Direction of Force from its Components
2. If the components of a certain
force
F,
referred
to
the
rectangular coordinate system,
are
Fx = - 315 N
and Fy = 245 N, find the
magnitude and direction of F.
Figure 2 Magnitude and Direction of Force F
SOLUTION 2: With the given components of F, use the following equations to solve
𝑀𝑎𝑔𝑛𝑖𝑡𝑢𝑑𝑒: 𝐹 =
2
𝐹𝑥
2
𝐹𝑦
+ 𝐹𝑦 and 𝐷𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛: α = 𝑎𝑟𝑐𝑡𝑎𝑛|| 𝐹𝑥 ||
Therefore, F = 399.061 N and α = 37.875°
ENS 161 STATICS of RIGID BODIES
Chapter 1: Statics of Particles
2nd Semester A.Y. 2024 - 2025
Sample Problem: Analytical Method of Solving for the Resultant of a Force System
3. Given the four concurrent forces
in the figure. Solve for the
magnitude and direction of the
resultant.
Figure 3 Magnitude and Direction of Resultant
SOLUTION 3: To find the components of the resultant via analytical method, use the following,
𝑅𝑥 = ∑ 𝐹𝑥 ;
𝑅𝑦 = ∑ 𝐹𝑦 ;
𝑅 =
𝑅𝑥 = ∑ 𝐹𝑥 = 𝐹1𝑥 + 𝐹2𝑥 − 𝐹3𝑥 − 𝐹4𝑥
𝑅𝑥 = 75 + 70
3
13
− 85
1
2
2
𝑅𝑥
2
+ 𝑅𝑦 ;
𝑅𝑦 = ∑ 𝐹𝑦 = 𝐹1𝑦 + 𝐹2𝑦 + 𝐹3𝑦 − 𝐹4𝑦
− 100𝑐𝑜𝑠60
𝑅𝑥 = 23. 139 𝑁 → (store exact value to
𝑅𝑦 = 0 + 70
Then, substitute stored values to the following equations,
2
𝑅𝑥
2
13
+ 85
1
2
− 100𝑠𝑖𝑛60
𝑅𝑥 = 12. 331 𝑁 ↑ (store exact value to B)
A)
𝑅 =
𝑅𝑦
θ = 𝑎𝑟𝑐𝑡𝑎𝑛|| 𝑅𝑥 ||
2
𝑅𝑦
+ 𝑅𝑦 and θ = 𝑎𝑟𝑐𝑡𝑎𝑛|| 𝑅𝑥 ||
Therefore, R = 26.230 N and θ = 28.052°
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