Rotational Equilibrium: The sum of all moments about any point must also be zero
$$\sum M = 0$$
In engineering mechanics, a non-concurrent force system refers to a group of forces whose lines of action do not intersect at a single point. These forces may act on different parts of a rigid body and can include forces and moments (torques). To achieve equilibrium in such systems, a body must satisfy both conditions above.
General Resultant of Non-concurrent Force Systems:
a. Calculate the location of the resultant from point D. b. Prove that the resultant force intersects point E (as indicated below) c. Determine the distance dx and dy from point D to the resultant force and indicate the location relative to D. d. Determine the distance from point B to the resultant force. e. Resolve the system into a resultant-couple system.
See images:
Below is a scaled drawing of the figure (modeled using AutoCad).We can obtain the moment caused by the resultant force about point D so that we can transfer the resultant so that it intersects point D. The moment (couple) will be applied at point D for clarity. Finally, we have this resultant-couple system that replaces the system above.
Problem:
For the beam shown, beam ABC is supporting beam CD.
Which of the following most nearly gives the reaction at C in kN.
A. 250
B. 150
C. 100
D. 200
Which of the following most nearly gives the reaction at D in kN.
A. 250
B. 150
C. 100
D. 200
Which of the following most nearly gives the value of "x" if the reaction at B is 380 N.
A. 1
B. 1.5
C. 2
D. 3
See images:
Problem:
The uniform 30kg bar with end rollers is supported by the horizontal and vertical surfaces and the wire AC.
a. Determine the tension in the wire.
b. Determine the reaction at A.
c. Determine the moment at B.
See images:
Problem:
The uniform rod weighs 420N and has its center of gravity at G as shown.
a. What is the reaction at B?
b. What is the tension in the cable?
c. What is the reaction at A?
See images:
Problem:
A three-legged stool is subjected to the load P as shown. If P=200N, compute the following:
a. Reaction under leg A
b. Reaction under leg B
c. Reaction under leg C
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Problem:
For the figure shown, compute the reaction at the hinged support at A if the weight of the cylinder is 1000N.
See images:
Problem:
For the figure shown, calculate the angle of tilt, θ, so that the contact force at B will be one-half that of A for the smooth cylinder.
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Problem: Walkway from a Pier to a Float
To accomodate the rise and fall of tide, a walkway from a pier to a float is supported by two rollers as shown. The center of mass of the 300-kg walkway is at G.
a. Calculate the force under the roller at A in Newtons.
b. Calculate the tension T in the horizontal cable which is attached to the cleat in Newtons.
c. Calculate the force in the roller at B in Newtons.
See images:
Problem: Cylinder Placed between a Rigid Member
The cylinder below has a mass of 200kg. Determine the following:
a. The normal reaction at D in Newtons.
b. The normal reaction at the wall at C in Newtons.
c. The tension in the cable at D in Newtons.
See images:
FBD of cylinder
$$
W = 200(9.81) = 1962\ \text{N}, \qquad r = 1\ \text{m}, \qquad h = 3\ \text{m}
$$
$$
\theta = 30^\circ, \qquad a = 90^\circ - 30^\circ = 60^\circ
$$
All moment equations are taken about point A, the hinge support, so the unknown hinge reactions at A are eliminated. The normal reaction used depends on the chosen FBD: the FBD considering the contact at D uses $N_D$ and does not include $N_C$, while the external FBD uses the normal reaction at C and does not include $N_D$. Since $N_D$ is the resultant effect of the cylinder's weight $W$ and the wall reaction $N_C$, $N_C$ is not written separately when $N_D$ is already being considered.
A 5m simply supported beam has a 10kN force applied at midspan, inclined 30° from the vertical toward the right. Determine the support reactions if A is a pin and B is a roller.
Resolve the inclined force first. The roller at B has only a vertical reaction.
The weight must resist the vertical (uplift) component of the resultant with the required factor of safety: $ W_{min} = FS \times \Sigma F_y = 1.3 \times 8.87 $ $ \boxed{W_{min} = 11.53 \text{ kN}} $
Question Bank: q2
PSAD - Statics / Equilibrium of Nonconcurrent Force Systems / Engr. Janclyde Espinosa (Clidez)
A boom supports a 100N load. If the tension in the cable is 200N,
Determine the angle α (in degrees)
21.47
20.92
23.45
19.87
Determine the angle θ (in degrees)
42.94
41.84
46.9
39.74
Determine the force in the boom in Newtons.
254.26 (C)
254.26 (T)
264.52 (C)
264.52 (T)
Solution pending in psadquestions/q2.json.
Question Bank: q3
PSAD - Statics / Equilibrium of Nonconcurrent Force Systems / Engr. Janclyde Espinosa (Clidez)
CE Board Nov. 2024 For the truss shown, members AD and BC are flexible cables.
CE Board November 2024 Two cylinders are loaded as shown and are connected by a rigid curved rod parallel to the smooth cylindrical surface. Neglect the diameter of the cylinders. W1=750N, W2=420N ϴ=90º, R=6m
If the surfaces are frictionless, what is α to maintain equilibrium?
60.75
50.17
45.14
76.12
If the coefficient of friction is μs=0.2 in the contacting surfaces, what is α to prevent rotation counterclockwise?
38.66
56.88
49.44
50.17
If the coefficient of friction is μs=0.015 in the contacting surfaces, which of the following gives α to impend motion?
For the assembly shown, beam AB weighs 100lbs. It is loaded with a force P=200lbs. A frictionless pulley with radius "r" weighs 80lbs and is attached at the end of the beam at point B, supported by two ropes inclined at angles x and y at points C and D. If a=2ft, b=3ft, angles x & y are 30º and 40º, respectively, and r=6 in.,
CE Board Nov 2011 and April 2025 A load W=30kN is lifted by a boom BCD making an angle, α=60º from the vertical axis as shown. Neglect the weight of the boom.
Determine the angle β between the cables AC and AD, in degrees. Note: Point C is midway points B and D.
A cylindrical rum 2m in radius is held by a rigid bar AB hinged at A and a flexible wire cable BC as shown. The drum weighs 1500N and strut AB has a length of 10m. Neglect friction in all contact surfaces.
Which of the following gives the normal reaction at D in Newtons.
3000
2000
3500
2500
Which of the following gives the vertical reaction at A in Newtons.
146.8
185.3
125.9
564.2
Which of the following gives the tension in cable BC in Newtons.
Compute the horizontal and vertical components of the support reactions of the compound beam shown. Each support is labeled as A, B, C, and D, respectively.
What is the horizontal reaction at A?
0
10
20
15
What is the vertical reaction at A?
20
10
15
40
What is the vertical reaction of the internal roller at B?
If the coefficient of static friction at contact points A and B in the figure below is μs=0.25, the weight of the spool is 70kg, and R1=0.6m; R2=0.9m, determine the following:
Maximum force P that can be applied without causing the spool to move.
210.214
448.457
112.114
103.48
Normal Reaction at B.
448.457
210.214
112.114
103.48
Normal Reaction at A.
112.114
448.457
210.214
103.48
Solution pending in psadquestions/q323.json.
Question Bank: q527
PSAD - Statics / Resultant/Equilibrium of Nonconcurrent Force Systems / Engr. Janclyde Espinosa (Clidez)
For the system shown:
Calculate the location of the resultant from point D.
0.832m left of D
0.832m right of D
1.5m left of D
1.5m right of D
Determine the horizontal distance from the resultant to point D and indicate the direction.
1.5m left of D
1.5m right of D
0.832m left of D
0.832m right of D
Determine the vertical distance from the resultant to point D and indicate the direction.
1m above D
1m below D
1.5m above D
1.5m below D
Determine the distance from point B to the resultant force.
0.55m
0.66m
0.44m
0.33m
Solution pending in psadquestions/q527.json.
Question Bank: q528
PSAD - Statics / Resultant/Equilibrium of Nonconcurrent Force Systems / Engr. Deguma
The uniform 30kg bar with end rollers is supported by the horizontal and vertical surfaces and the wire AC.
Determine the tension in the wire in Newtons.
295.34
235.44
196.20
73.55
Determine the reaction at A in Newtons.
73.55
76.48
83.42
90.54
Determine the moment at B in Newton-meter.
235.44
295.34
196.20
194.72
Solution pending in psadquestions/q528.json.
Question Bank: q530
PSAD - Statics / Resultant/Equilibrium of Nonconcurrent Force Systems / Engr. Deguma
A three-legged stool is subjected to the load P as shown. If P=200N, compute the following:
Reaction under leg A in Newtons
106.67
46.67
143.33
56.67
Reaction under leg B in Newtons
46.67
106.67
143.33
56.67
Reaction under leg C in Newtons
46.67
106.67
143.33
56.67
Solution pending in psadquestions/q530.json.
Question Bank: q531
PSAD - Statics / Resultant/Equilibrium of Nonconcurrent Force Systems / Engr. Deguma
The weight of the cylinder in the figure shown is 1000N.
Compute the normal force between the cylinder and bar AB.
1666.67
833.33
1000.00
1118.03
Compute the tension in cable BC.
833.33
1118.03
1000.00
1666.67
Compute the reaction at the hinged support at A.
1118.03
1666.67
1000.00
833.33
Solution pending in psadquestions/q531.json.
Question Bank: q532
PSAD - Statics / Resultant/Equilibrium of Nonconcurrent Force Systems / Engr. Deguma
For the figure shown, calculate the angle of tilt, θ, so that the contact force at B will be one-half that of A for the smooth cylinder.
Forces P1 = 1.8 kN (30°), P2 = 0.9 kN (vertical), and P3 = 0.45 kN (45°) act on a beam.
Determine the magnitude of the resultant.
2.45 kN
3.15 kN
2.12 kN
1.24 kN
From the beam system in Question 34, calculate the vertical reaction at support B, assuming the beam is 2 m long and forces act at the midpoint.
1.07 kN
2.12 kN
0.90 kN
1.24 kN
From the beam system in Question 34, determine the horizontal reaction at support B, assuming B is a roller support.
0 kN
1.24 kN
1.56 kN
0.45 kN
Part 1.
Resolve the inclined forces into components. Taking the horizontal components of $P_1$ and $P_3$ in opposite directions: $R_x = 1.8\cos 30^\circ - 0.45\cos 45^\circ = 1.241$ kN $R_y = 0.9 + 1.8\sin 30^\circ + 0.45\sin 45^\circ = 2.118$ kN $R = \sqrt{R_x^2 + R_y^2} = \sqrt{1.241^2 + 2.118^2}$ $\boxed{R = 2.45\text{ kN}}$
Part 2.
The total downward vertical component is: $R_y = 0.9 + 1.8\sin 30^\circ + 0.45\sin 45^\circ = 2.118$ kN Since the forces act at the midpoint of a simply supported 2 m beam, the vertical reactions are equal: $B_y = \frac{R_y}{2} = \frac{2.118}{2}$ $\boxed{B_y = 1.07\text{ kN}}$
Part 3.
A roller support cannot resist horizontal force. Therefore the horizontal reaction at support B is zero regardless of the applied horizontal components. $\boxed{B_x = 0\text{ kN}}$
A storage tank (W = 2000 N, r = 1.0 m) is pushed over a 0.25 m step.
Find the horizontal force F required to just break contact at point A.
1732 N
2000 N
2646 N
1500 N
From Question 43, find the magnitude of the reaction at point B at the instant contact at A is broken.
2646 N
2000 N
1732 N
3732 N
Part 1.
At impending motion over the step, contact at A is just broken and the tank tends to rotate about point B. Taking moments about B eliminates the reaction at B. $F(1.00) = 2000(0.866)$ $F = 1732$ N $\boxed{F = 1732\text{ N}}$
Part 2.
At the instant contact at A is broken, equilibrium requires the reaction at B to balance the horizontal push and the weight components: $R_B = \sqrt{F^2 + W^2}$ $R_B = \sqrt{1732^2 + 2000^2}$ $\boxed{R_B = 2646\text{ N}}$