Equilibrium and Friction Notes FE Sem 1 Mechanics | Mumbai Univ

A body is in equilibrium when the resultant of all forces and couples on it is zero, and this module turns that idea into working equations. You will draw free body diagrams, apply Lami's theorem and the three equations of equilibrium, and find support reactions of simple, overhanging and cantilever beams under point loads, uniformly distributed and uniformly varying loads. The last part introduces dry friction: its laws, angle of friction, cone of friction and angle of repose, applied to bodies on horizontal and inclined planes. This module carries the largest teaching hours in the course and usually the most marks.

Summary

Conditions of equilibrium: for coplanar concurrent forces the conditions are Sum Fx = 0 and Sum Fy = 0 (two equations, two unknowns); Lami's theorem can be used when exactly three concurrent forces act. For coplanar parallel forces, Sum F = 0 and Sum M = 0 about any point. For a general coplanar force system, Sum Fx = 0, Sum Fy = 0 and Sum M = 0 about any point, so at most three unknowns can be found. Applied couples enter only the moment equation; a couple can be balanced only by another couple.

Equilibrium of rigid bodies and free body diagrams: a free body diagram (FBD) isolates the body and shows all external forces on it, namely applied loads, self weight, and the reactions that replace every support or contact. Smooth surfaces give a reaction normal to the surface, cables give tension along the cable away from the body, hinges give two unknown components, rollers give one reaction normal to the rolling surface, and fixed supports give two components and a moment. Two-force members carry forces along the line joining the two points; for a three-force body the three forces must be concurrent or parallel.

Types of beams and supports: beams are simply supported, cantilever, overhanging, fixed, propped cantilever and continuous; a compound beam is two or more beams joined together (problems on internal hinges are excluded in this syllabus). Supports are roller (one reaction perpendicular to the surface), hinge or pin (horizontal and vertical reactions) and fixed (horizontal, vertical reactions and a fixing moment). Loads may be point loads, inclined loads, uniformly distributed loads (UDL), uniformly varying loads (UVL) and applied couples.

Determination of support reactions: replace each distributed load by its equivalent point load. A UDL of intensity w over length L equals wL acting at its midpoint; a UVL from zero to w over length L equals wL/2 acting at L/3 from the larger end. Resolve inclined loads into components. Then apply Sum M = 0 about one support to find the other reaction, Sum Fy = 0 for the remaining vertical reaction, and Sum Fx = 0 for the horizontal reaction at a hinge. A couple applied anywhere on the beam enters the moment equation with its full value, independent of its position.

Friction and its laws: friction is the tangential resisting force at the contact surface opposing relative motion or its tendency. Static friction adjusts up to a limiting value F = mu N at impending motion; after motion begins the kinetic friction is slightly less. Laws of dry (Coulomb) friction: friction acts opposite to the tendency of motion; limiting friction is proportional to the normal reaction; it is independent of the area of contact; it depends on the nature of the surfaces; and kinetic friction is nearly independent of velocity for moderate speeds.

Angle of friction, cone of friction and angle of repose: the angle of friction phi is the angle between the normal reaction and the resultant reaction at limiting friction, tan phi = mu. Revolving this resultant about the normal generates the cone of friction with semi-vertex angle phi; the body stays at rest as long as the resultant reaction lies within the cone. The angle of repose is the maximum inclination of a plane on which a body rests without sliding under its own weight; it equals the angle of friction.

Equilibrium of bodies on horizontal and inclined planes: draw the FBD with weight, normal reaction, friction (opposing the impending motion) and the applied force. On a horizontal plane with a pull P at angle theta above the horizontal, N = W - P sin theta and P cos theta = mu N at impending motion. On an inclined plane of angle alpha, resolve along and perpendicular to the plane: N = W cos alpha (plus any perpendicular component of P) and the force along the plane must overcome W sin alpha plus or minus mu N depending on whether motion is impending up or down. Ladder problems use friction at the floor and wall together with the three equilibrium equations.

Mind map

Procedure to find support reactions of a beam

Important definitions

Formulas

Key points to remember

• Draw a separate free body diagram for each body; most errors come from missing or wrongly directed reactions.
• Lami's theorem works only for exactly three concurrent forces in equilibrium and uses the angles between the forces.
• A roller reaction is perpendicular to the surface on which the roller rests, not necessarily vertical.
• Take moments about a hinge first because its two unknown components drop out.
• A UVL acts at L/3 from the larger ordinate and 2L/3 from the zero end.
• An applied couple adds directly to the moment equation and does not depend on where it acts.
• Friction always opposes the impending relative motion; reverse its direction for the 'just prevent slipping down' case.
• Friction F equals mu N only at impending motion; otherwise F is found from equilibrium and must be less than mu N.
• Limiting friction is independent of the area of contact and depends on the normal reaction, not on the weight directly.
• Angle of repose equals angle of friction, so a block slides down unaided if the incline is steeper than phi.

Solved examples

Example 1. A smooth cylinder of weight 500 N rests between two smooth planes inclined at 30 degrees and 60 degrees to the horizontal (a V-groove). Find the reactions of the planes.

Reactions are normal to the planes and pass through the centre. R1 (from the 30 degree plane) makes 30 degrees with the vertical; R2 (from the 60 degree plane) makes 60 degrees with the vertical on the other side. Angles between forces: R1 and R2 = 30 + 60 = 90 degrees; W and R1 = 180 - 30 = 150 degrees; W and R2 = 180 - 60 = 120 degrees. Lami's theorem: R1/sin 120 = R2/sin 150 = 500/sin 90. R1 = 500 x 0.866 = 433.0 N, R2 = 500 x 0.5 = 250 N. Check horizontal: 433 sin 30 = 216.5 = 250 sin 60. Vertical: 433 cos 30 + 250 cos 60 = 375 + 125 = 500 N.

Example 2. A simply supported beam AB of span 6 m has a hinge at A and a roller at B. It carries a point load of 20 kN at 2 m from A and a UDL of 10 kN/m over the last 3 m towards B. Find the support reactions.

UDL equivalent = 10 x 3 = 30 kN at 4.5 m from A. Sum M_A = 0 (anticlockwise positive): R_B(6) - 20(2) - 30(4.5) = 0, so R_B = (40 + 135)/6 = 29.17 kN upward. Sum Fy = 0: R_A + 29.17 - 20 - 30 = 0, R_A = 20.83 kN upward. Sum Fx = 0 gives horizontal reaction at A = 0. Check about B: 20.83(6) - 20(4) - 30(1.5) = 125 - 80 - 45 = 0.

Example 3. An overhanging beam ABC is hinged at A and rests on a roller at C, 6 m from A; the overhang CB is 2 m. It carries an inclined load of 20 kN at 3 m from A acting at 60 degrees to the beam (pointing down and to the right) and a vertical load of 10 kN at the free end B. Find the reactions.

Components of the inclined load: vertical 20 sin 60 = 17.32 kN down, horizontal 20 cos 60 = 10 kN to the right. Sum M_A = 0: R_C(6) - 17.32(3) - 10(8) = 0, R_C = (51.96 + 80)/6 = 21.99 kN upward. Sum Fy = 0: R_Ay = 17.32 + 10 - 21.99 = 5.33 kN upward. Sum Fx = 0: R_Ax = 10 kN to the left. Resultant at A: R_A = square root of (10^2 + 5.33^2) = 11.33 kN at arctan(5.33/10) = 28.05 degrees above the horizontal (pointing left and up).

Example 4. A cantilever AB of length 4 m is fixed at A. It carries a uniformly varying load from 12 kN/m at A to zero at B and a point load of 10 kN at B. Find the reactions at the fixed support.

UVL equivalent = 0.5 x 12 x 4 = 24 kN, acting at L/3 = 4/3 = 1.333 m from A (the larger end). Sum Fy = 0: R_A = 24 + 10 = 34 kN upward. Sum Fx = 0: H_A = 0. Sum M_A = 0: M_A - 24(1.333) - 10(4) = 0, M_A = 32 + 40 = 72 kN m, acting anticlockwise on the beam (fixing moment at A).

Example 5. A block weighing 500 N rests on a horizontal floor (mu = 0.3). Find the force P, inclined at 30 degrees above the horizontal, required just to move the block.

FBD: W = 500 N down, P at 30 degrees, N up, friction F = mu N opposing motion. Sum Fy = 0: N + P sin 30 = 500, N = 500 - 0.5P. Sum Fx = 0 at impending motion: P cos 30 = 0.3N = 0.3(500 - 0.5P). 0.866P + 0.15P = 150, P = 150/1.016 = 147.63 N. Then N = 426.18 N and F = 127.85 N.

Example 6. A block of weight 1000 N is placed on a plane inclined at 30 degrees to the horizontal, mu = 0.25. Find (a) the force parallel to the plane to just move it up, (b) the force parallel to the plane to just prevent it sliding down, (c) the horizontal force to just push it up.

N = W cos 30 = 866.03 N, limiting friction = 0.25 x 866.03 = 216.51 N; W sin 30 = 500 N. (a) Motion up, friction acts down the plane: P = 500 + 216.51 = 716.51 N. (b) Since tan 30 = 0.577 > 0.25 the block tends to slide down; friction acts up the plane: P = 500 - 216.51 = 283.49 N. (c) phi = arctan 0.25 = 14.04 degrees. With a horizontal force P, N = W cos 30 + P sin 30 and P cos 30 = W sin 30 + mu N, which gives P = W tan(30 + 14.04) = 1000 tan 44.04 degrees = 966.9 N.

Example 7. A uniform ladder of weight 200 N rests against a vertical wall with coefficient of friction 0.2 and on a floor with coefficient of friction 0.3. Find the least angle the ladder can make with the floor without slipping and the reactions.

Let the ladder length be L at angle theta with the floor, about to slip with its foot moving away from the wall. Floor: N1 up, F1 = 0.3N1 towards the wall. Wall: N2 horizontal, F2 = 0.2N2 upward. Sum Fx = 0: N2 = 0.3N1. Sum Fy = 0: N1 + 0.2N2 = 200, so N1(1 + 0.06) = 200, N1 = 188.68 N, N2 = F1 = 56.60 N, F2 = 11.32 N. Moments about the foot: 200(L/2) cos theta = N2 L sin theta + F2 L cos theta. 100 cos theta - 11.32 cos theta = 56.60 sin theta, tan theta = 88.68/56.60 = 1.5667, theta = 57.45 degrees. The ladder will not slip if it makes at least 57.45 degrees with the floor.

Important questions and answers

Q1. State and prove Lami's theorem. (4 marks)

Statement: if three coplanar concurrent forces P, Q and R acting at a point are in equilibrium, each force is proportional to the sine of the angle between the other two: P/sin alpha = Q/sin beta = R/sin gamma. Proof: since the forces are in equilibrium they can be represented by the sides of a closed triangle taken in order. The interior angles of this triangle are (180 - alpha), (180 - beta), (180 - gamma), where alpha, beta and gamma are the angles between the forces. By the sine rule, P/sin(180 - alpha) = Q/sin(180 - beta) = R/sin(180 - gamma), and since sin(180 - x) = sin x, the theorem follows.

Q2. State the conditions of equilibrium for concurrent, parallel and general coplanar force systems. (3 marks)

Concurrent: Sum Fx = 0 and Sum Fy = 0. Parallel: Sum F = 0 (along the direction of forces) and Sum M = 0 about any point. General: Sum Fx = 0, Sum Fy = 0 and Sum M = 0 about any point. In each case the resultant force is zero and, for non-concurrent systems, the resultant couple is also zero.

Q3. Explain types of supports with their reactions and sketches. (4 marks)

Roller support: allows rotation and movement along the surface; provides one reaction perpendicular to the surface. Hinge or pin support: allows rotation but no translation; provides horizontal and vertical reaction components. Fixed support: prevents rotation and translation; provides horizontal and vertical reactions and a fixing moment. Simple support: a knife edge giving a reaction normal to the beam. Sketches show the symbol and the reaction arrows for each.

Q4. Explain types of beams and types of loads. (4 marks)

Beams: simply supported (supports at both ends), cantilever (fixed at one end, free at the other), overhanging (extends beyond a support), fixed (both ends fixed), propped cantilever, continuous (more than two supports) and compound beams (segments joined together). Loads: point or concentrated load, inclined load, uniformly distributed load of w per metre, uniformly varying load (triangular or trapezoidal) and applied couple or moment.

Q5. State the laws of dry friction. (3 marks)

1. Friction acts tangential to the contact surfaces and opposite to the direction of impending or actual relative motion. 2. Limiting friction is directly proportional to the normal reaction, F = mu N. 3. Limiting friction is independent of the area of contact for a given normal reaction. 4. It depends on the nature and condition of the surfaces in contact. 5. Kinetic friction is slightly less than limiting friction and is nearly independent of velocity for moderate speeds.

Q6. Define angle of friction, angle of repose and cone of friction, and show that angle of repose equals angle of friction. (5 marks)

Angle of friction phi is the angle between the normal reaction N and the resultant R of N and limiting friction F, so tan phi = F/N = mu. Angle of repose is the maximum inclination alpha of a plane on which a body rests without sliding under its own weight. Cone of friction is generated by revolving R about N, with semi-vertex angle phi. Proof: for a block of weight W on a plane at the angle of repose alpha, at impending sliding: along the plane W sin alpha = F = mu N, perpendicular N = W cos alpha. So tan alpha = mu = tan phi, giving alpha = phi.

Q7. A 2 m long horizontal beam AB is simply supported and carries a clockwise couple of 6 kN m at its midpoint. Find the reactions. (3 marks)

There is no net vertical load, so the reactions form a couple opposing the applied couple. Sum M_A = 0 (anticlockwise positive): R_B(2) - 6 = 0, R_B = 3 kN upward. Sum Fy = 0: R_A = -3 kN, i.e. 3 kN downward. The reactions do not depend on where the couple acts.

Q8. Show that the least force required to drag a body on a rough horizontal plane is W sin phi and acts at angle phi to the horizontal. (4 marks)

Let P act at angle theta above horizontal. N = W - P sin theta, and at impending motion P cos theta = mu(W - P sin theta), so P = mu W/(cos theta + mu sin theta). Putting mu = tan phi: P = W sin phi/(cos theta cos phi + sin theta sin phi) = W sin phi/cos(theta - phi). P is least when cos(theta - phi) = 1, i.e. theta = phi, giving P_min = W sin phi.

Q9. Explain the free body diagram with an example and state its importance. (3 marks)

An FBD is a sketch of a body detached from all surroundings, showing every external force: weights, applied loads and reactions replacing supports and contacts. Example: a ladder resting on a floor and wall shows its weight at the midpoint, normal and friction forces at the foot and at the top. It is essential because the equilibrium equations can only be written correctly once all forces with their directions are identified.

Q10. Explain a two-force member and a three-force body with their equilibrium conditions. (3 marks)

A two-force member has forces applied at only two points; for equilibrium the two forces must be equal, opposite and collinear along the line joining the points, as in a link or strut. A three-force body has forces at three points; for equilibrium the three forces must be either concurrent or parallel, which allows Lami's theorem or a graphical triangle to be used.

Exam tips

• Draw the FBD first and label every reaction; marks are given for it separately.
• Write the moment equation about the hinge first, then the force equations.
• Clearly show the equivalent point load and its position for each UDL and UVL.
• Check beam reactions by a moment equation about another point.
• In friction problems, state which way motion is impending and draw friction opposite to it.
• For inclined planes, use axes along and perpendicular to the plane to simplify equations.
• Problems on internal hinges are excluded from the syllabus, so do not spend time on them.

FAQs

When can I use Lami's theorem?

Only when exactly three coplanar concurrent forces keep a body in equilibrium. With more forces use Sum Fx = 0 and Sum Fy = 0.

What if a reaction comes out negative?

It means the actual direction is opposite to the one assumed. Keep the sign while substituting in later equations and state the true direction in the final answer.

Is friction always equal to mu N?

No. F = mu N only at impending motion (limiting state). In ordinary rest, friction equals whatever is required for equilibrium, up to mu N.

Is the area of contact relevant in friction problems?

No. By the laws of dry friction, limiting friction is independent of the area of contact for a given normal reaction.

These notes follow the University of Mumbai first-year engineering syllabus (NEP 2020, Engineering Mechanics) and were prepared with AI help and checked.