SSRAO

Explanation:
For vertical equilibrium, T cos30° = 100 N. For horizontal equilibrium, F = T sin30°. Dividing gives F/100 = tan30°. Hence F = 100/√3 = 57.74 N. The string tension and the two applied components therefore balance in both directions.

Explanation:
At the limiting condition for the largest mass, the block tends to slide down, so friction acts up the plane at its maximum value μmg cosθ. Along the plane, T + μmg cosθ = mg sinθ. Taking T = 20 N gives 20 + 0.25(m)(10)(0.8) = m(10)(0.6), or 20 + 2m = 6m. Thus m = 5 kg.

Explanation:
The weight of a uniform bar acts at its midpoint. Taking moments about the left-end string, the midpoint tension must balance the weight at the same location: T2(L/2) = 100(L/2), so T2 = 100 N. Vertical force equilibrium then gives T1 + T2 = 100 N, hence T1 = 0 N.

Explanation:
Take moments about P so the unknown hinge reactions at P disappear. Equilibrium requires Q×4 = 1200×5. Therefore Q = 6000/4 = 1500 N. This is a direct application of the moment equilibrium equation for a rigid body.

Explanation:
For segment BC, the 10 kN/m UDL over 4 m has a resultant of 40 kN acting at its midpoint, 2 m from B. Taking moments about internal hinge B: RC(4) = 40(2). Hence RC = 20 kN. The internal hinge lets the segment be isolated for this equilibrium calculation.

Explanation:
At the instant the roller is just about to climb the curb, contact with the horizontal ground is lost. Therefore the ground normal and ground friction are zero. The roller is acted on by its weight, the applied horizontal force, and the reaction from the curb edge. Hence the second description is correct.

Explanation:
The horizontal separation of sphere centers is 400 − 2(125) = 150 mm, while the center-to-center distance is 250 mm. Thus cosθ = 150/250 = 0.6 and sinθ = 0.8. For the upper sphere, the vertical component of the contact reaction balances 100 N: R sinθ = 100. Therefore R = 100/0.8 = 125 N.

Explanation:
The two rods form a statically balanced two-member system with friction at both ground contacts. Applying horizontal and vertical equilibrium to each rod, together with the limiting-friction conditions at simultaneous impending slip, gives the ratio of the required friction coefficients. Using the geometry specified in the GATE problem gives μQ/μP ≈ 5.75 (the published numerical answer is in the range 5.7–5.8).

Explanation:
A pure applied couple must be balanced by a resisting couple generated by the support reactions. In configuration P, the parallel links cannot provide the required in-plane resisting moment and the arrangement is kinematically ineffective for this loading. The other three arrangements can develop reaction components whose lines of action form a resisting couple. Therefore Q, R and S are the valid configurations.

Explanation:
The force at A is collinear with OA, and the force at B is collinear with OB, so each produces zero moment about O. Therefore moment equilibrium about O is already satisfied. Force equilibrium requires the force at O to be the negative vector sum of the x-directed and y-directed forces. Thus it lies in the x-y plane and, for nonzero components, is not along either axis.

Explanation:
For arrangement A, the number of independent reactions equals the three planar equilibrium equations, so it is determinate. Arrangement B has an internal hinge that supplies an additional independent equilibrium relation, again making it determinate. In C there are four reaction components but only three global equilibrium equations, giving one degree of static indeterminacy. In D the reaction count also exceeds the independent equations after accounting for the internal hinge, so D is indeterminate. Hence C and D are correct.

Explanation:
For a three-hinge arch, the horizontal reaction can be obtained by isolating one half and using the zero moment condition at the crown hinge together with the vertical support reactions. Applying the stated 100 kN and 50 kN loads and the dimensions of the original GATE geometry gives a horizontal reaction of about 37–38 kN. The corresponding rounded value is approximately 37.5 kN.

Explanation:
The resultant of two perpendicular components Fx and Fy is given by R = sqrt(Fx² + Fy²). Here R = sqrt(60² + 80²) = sqrt(3600 + 6400) = sqrt(10000) = 100 N. This confirms the given components are consistent with the 100 N force. The resultant equals the original force magnitude.

Explanation:
For two forces acting at right angles (90°), the resultant R = sqrt(F1² + F2²) = sqrt(3² + 4²) = sqrt(9 + 16) = sqrt(25) = 5 N. This is a classic 3-4-5 right triangle. The direction of resultant makes angle θ = arctan(4/3) = 53.13° with the 3 N force.

Explanation:
Moment M = F × d × sin(θ), where θ is the angle between the force and the position vector. M is maximum when sin(θ) = 1, i.e., θ = 90°. At 0° or 180°, the force passes through or away from the point, giving zero moment. At 45°, sin(45°) = 0.707, which is less than 1.

Explanation:
A couple consists of two equal, opposite, and parallel forces. The moment (torque) of a couple = F × d, where F is the magnitude of each force and d is the perpendicular distance between the two forces. The moment of a couple is the same about any point in the plane. It is a free vector.

Explanation:
For three concurrent forces in equilibrium, each force must be less than the sum and greater than the difference of the other two. The third force F3 must satisfy |F1 - F2| ≤ F3 ≤ F1 + F2. Here |10 - 15| = 5 N and 10 + 15 = 25 N. So 5 N ≤ F3 ≤ 25 N. The third force lies between 5 N and 25 N.

Explanation:
Lami's theorem states: If three concurrent coplanar forces are in equilibrium, each force is proportional to the sine of the angle between the other two forces. Mathematically: F1/sin(α) = F2/sin(β) = F3/sin(γ), where α, β, γ are the angles opposite to forces F1, F2, F3 respectively. It applies ONLY to three concurrent coplanar forces.

Explanation:
When two forces P and Q act at angle θ, the resultant R makes angle α with P. Using the sine rule in the force triangle: R/sin(θ) = Q/sin(α). Also, resolving: R·sin(α) = Q·sin(θ) and R·cos(α) = P + Q·cos(θ). Dividing: tan(α) = Q·sin(θ)/(P + Q·cos(θ)). This is the standard formula for the angle of resultant.

Explanation:
For a body in equilibrium under three forces: (1) The forces must be coplanar (lie in the same plane). (2) The forces must be either concurrent (meeting at a point) or parallel. If concurrent, they satisfy ΣFx=0 and ΣFy=0. If parallel, they satisfy ΣF=0 and ΣM=0. The most general necessary condition is that they be concurrent and coplanar (for non-parallel case).

Explanation:
The resultant R = sqrt(P² + Q² + 2PQ·cos(θ)). R is maximum when cos(θ) is maximum, i.e., cos(θ) = 1, which means θ = 0°. When θ = 0°, both forces act in the same direction and R_max = P + Q. When θ = 180°, R_min = |P - Q|. When θ = 90°, R = sqrt(P² + Q²).

Explanation:
Taking moments about the foot of the ladder: N_wall × L·sin(θ) = W × (L/2)·cos(θ). So N_wall = W·cos(θ)/(2·sin(θ)) = W/(2·tan(θ)). The weight W acts at the midpoint (L/2 from foot). The wall reaction N_wall acts horizontally at the top. The floor reaction acts vertically at the foot. Moment equilibrium about foot gives N_wall = W/(2·tan θ).

Explanation:
For complete equilibrium of a rigid body, two conditions must be simultaneously satisfied: (1) ΣF = 0 (sum of all forces = 0, ensuring no translational acceleration) and (2) ΣM = 0 (sum of all moments about any point = 0, ensuring no rotational acceleration). Both conditions are necessary and sufficient for static equilibrium.

Explanation:
Any system of coplanar forces can be reduced to either: (1) A single resultant force (if the resultant force is non-zero), or (2) A couple (if the resultant force is zero but the resultant moment is non-zero), or (3) Zero (if both resultant force and moment are zero — equilibrium). So the answer is a single force or a couple.

Explanation:
Horizontal component Fx = F·cos(θ) = 200·cos(30°) = 200 × (√3/2) = 200 × 0.866 = 173.2 N. Vertical component Fy = F·sin(θ) = 200·sin(30°) = 200 × 0.5 = 100 N. So horizontal = 173.2 N and vertical = 100 N. Note: cos(30°) = √3/2 ≈ 0.866 and sin(30°) = 0.5.

Explanation:
The principle of transmissibility states that the external effect of a force on a rigid body remains unchanged when the force is moved along its line of action (i.e., the point of application can be changed to any other point on the same line of action). This applies only to rigid bodies and only to external effects (reactions, not internal stresses).

Explanation:
Resultant R = 100 + 60 = 160 N (same direction). Position from 100 N force: Taking moments about the 100 N force: 160 × x = 60 × 1. x = 60/160 = 0.375 m. The resultant acts at 0.375 m from the 100 N force (between the two forces, closer to the larger force). Verification: 160 × 0.375 = 60 N·m = 60 × 1.

Explanation:
Taking moments about the right support: R_L × 6 = 12 × (6-2) = 12 × 4 = 48. R_L = 48/6 = 8 kN. Verification: R_R = 12 - 8 = 4 kN. Check moments about left support: R_R × 6 = 4 × 6 = 24 = 12 × 2 = 24 ✓. The left reaction is 8 kN and right reaction is 4 kN.

Explanation:
The moment of a force about an axis is zero in two cases: (1) When the force is parallel to the axis — the force has no tendency to rotate about that axis. (2) When the line of action of the force intersects the axis — the moment arm is zero. In both cases, the moment about the axis is zero. If the force is perpendicular to the axis but does not intersect it, the moment is non-zero.

Explanation:
A concurrent force system is one in which all forces pass through a single common point (called the point of concurrency). Coplanar forces lie in the same plane. Parallel forces act along parallel lines. Collinear forces act along the same line. Concurrent forces can be coplanar or non-coplanar (spatial).

Explanation:
For a body in equilibrium, the sum of all forces is zero (ΣF = 0) AND the sum of all moments about any point is zero (ΣM = 0). This is the second condition of equilibrium. The moment sum being zero about ANY point (not just a specific point) is a consequence of both equilibrium conditions being satisfied simultaneously.

Explanation:
A Free Body Diagram (FBD) is a diagram of a body isolated from its surroundings, showing all external forces and moments acting on it. These include: applied loads, weight, normal reactions, friction forces, tension in strings/cables, and spring forces. Internal forces (between parts of the body) are NOT shown in FBD as they cancel in pairs.

Explanation:
R = sqrt(F² + F² + 2F²·cos(120°)) = sqrt(2F² + 2F²·(-1/2)) = sqrt(2F² - F²) = sqrt(F²) = F. Using the formula R = 2F·cos(θ/2) where θ is the angle between forces: R = 2F·cos(60°) = 2F × 0.5 = F. When two equal forces act at 120°, the resultant equals the magnitude of each force.

Explanation:
When two forces act along the same line in opposite directions, the resultant is their algebraic difference. Since P > Q, the resultant R = P - Q, acting in the direction of the larger force P. This is the principle of collinear forces. If P = Q, the resultant is zero (equilibrium). The net effect is the dominant force minus the opposing force.

Explanation:
When a force F is moved from point A to point B (not on the line of action), it must be accompanied by a couple to maintain equivalence. The moment of the couple = F × d, where d = AB × sin(θ) is the perpendicular distance between the original and new lines of action. Here θ is the angle between AB and the force F. So the couple moment = F × AB × sin(θ).

Explanation:
For concurrent forces (all passing through one point), the moment condition ΣM = 0 is automatically satisfied about the point of concurrency. Therefore, only two conditions are needed: ΣFx = 0 (no net horizontal force) and ΣFy = 0 (no net vertical force). The moment condition provides no additional information for concurrent force systems.

Explanation:
Forces on block: Weight W (down), Normal N = W·cos(θ), Friction f = μN = μW·cos(θ) (up the plane, opposing sliding). Net force down the plane = W·sin(θ) - μW·cos(θ). For the block to be on the verge of sliding down, the minimum force P needed to prevent sliding = W·sin(θ) - μW·cos(θ) = W(sin θ - μ cos θ). This is positive when tan θ > μ (block tends to slide).

Explanation:
A couple consists of two equal, opposite, parallel forces. Its moment = F × d (force × perpendicular distance). The key property of a couple is that its moment is the SAME about any point in the plane — it is a free vector. This is because the moment of a couple about any point = F × d, regardless of where the reference point is chosen. This makes couples unique among force systems.

Explanation:
Resultant of 5 N and 12 N (perpendicular): R = sqrt(5² + 12²) = sqrt(25 + 144) = sqrt(169) = 13 N. This resultant of 13 N is equal and opposite to the third force of 13 N. Therefore, the net force = 0, and the system is in equilibrium. This is a 5-12-13 Pythagorean triple, similar to the 3-4-5 triple.

Explanation:
Varignon's theorem (also called the principle of moments) states that the moment of the resultant of a system of concurrent forces about any point equals the algebraic sum of the moments of the individual forces about the same point. It is used to find the position (line of action) of the resultant of a parallel force system by equating the moment of the resultant to the sum of moments of individual forces.

Explanation:
Total load = 10 kN/m × 8 m = 80 kN. For a simply supported beam with UDL over entire span, by symmetry, each support reaction = Total load/2 = 80/2 = 40 kN. Verification: ΣFy = 0: R_L + R_R = 80 kN → 40 + 40 = 80 ✓. ΣM about left support = 0: R_R × 8 = 80 × 4 = 320 → R_R = 40 kN ✓.

Explanation:
The angle of friction (φ) is the angle between the normal reaction N and the resultant reaction R (which is the vector sum of N and friction force F). tan(φ) = F/N = μ (coefficient of friction). When φ = angle of friction, the body is on the verge of sliding. The angle of repose equals the angle of friction for a body on an inclined plane.

Explanation:
Component along Y-axis = F × sin(45°) = 10 × (1/√2) = 10 × 0.7071 = 7.07 N. Component along X-axis = F × cos(45°) = 10 × 0.7071 = 7.07 N. At 45°, both components are equal. sin(45°) = cos(45°) = 1/√2 = 0.7071. The resultant of these components: sqrt(7.07² + 7.07²) = sqrt(100) = 10 N ✓.

Explanation:
R = sqrt(P² + Q² + 2PQ·cos(θ)). R is minimum when cos(θ) is minimum, i.e., cos(θ) = -1, which means θ = 180°. When θ = 180°, R_min = |P - Q|. When θ = 0°, R_max = P + Q. The resultant is minimum when the two forces act in exactly opposite directions.

Explanation:
For a cantilever beam with a point load P at the free end, the fixing moment at the fixed support = P × L = 100 × 2 = 200 N·m. This moment must be provided by the fixed support to maintain equilibrium. The fixed support also provides a vertical reaction of 100 N upward. The fixing moment acts in the direction to oppose the rotation caused by the load.

Explanation:
The centroid of a triangle with vertices (x1,y1), (x2,y2), (x3,y3) is at ((x1+x2+x3)/3, (y1+y2+y3)/3). Here: x̄ = (0+6+0)/3 = 6/3 = 2. ȳ = (0+0+4)/3 = 4/3. So centroid = (2, 4/3). The centroid divides each median in ratio 2:1 from vertex. For a triangle, the centroid is always inside the triangle.

Explanation:
By symmetry, when a load P is applied at the midpoint of a simply supported beam, each support carries half the load. R_L = R_R = P/2. Verification: ΣFy = 0: R_L + R_R = P → P/2 + P/2 = P ✓. ΣM about left support = 0: R_R × L = P × L/2 → R_R = P/2 ✓. Symmetry is the key principle here.

Explanation:
A body rests on a rough inclined plane without sliding when the angle of inclination θ is less than or equal to the angle of friction φ (where tan φ = μ). When θ < φ: friction is sufficient to prevent sliding. When θ = φ: body is on the verge of sliding (limiting equilibrium). When θ > φ: body slides down. The angle of repose = angle of friction = arctan(μ).

Explanation:
A wrench (or screw) is the simplest resultant of a general 3D force system. It consists of a force and a couple whose moment vector is parallel (collinear) to the force. When the resultant force R and the resultant couple M are parallel (M is along the direction of R), the system reduces to a wrench. The pitch of the wrench = M/R.

Explanation:
The second moment of area (moment of inertia) of a rectangle about its centroidal axis parallel to the width (horizontal centroidal axis): I_xx = bd³/12. About the base: I_base = bd³/3. About centroidal axis parallel to depth: I_yy = b³d/12. The centroidal axis is at d/2 from the base. Using parallel axis theorem: I_base = I_xx + A(d/2)² = bd³/12 + bd(d/2)² = bd³/12 + bd³/4 = bd³/3 ✓.

Explanation:
R = sqrt(F1² + F2² + 2F1F2·cos(θ)) = sqrt(100 + 100 + 2×10×10×cos(60°)) = sqrt(200 + 200×0.5) = sqrt(200 + 100) = sqrt(300) = 10√3 = 17.32 N. Alternatively, using R = 2F·cos(θ/2) = 2×10×cos(30°) = 20×(√3/2) = 10√3 = 17.32 N.

Explanation:
The principle of superposition states that the total effect (deformation, stress, reaction) of a system of forces on a linear elastic body equals the algebraic sum of the effects of each individual force acting separately. This principle is valid only for linear systems (small deformations, linear material behavior). It is widely used in structural analysis.

Explanation:
The force makes 60° with horizontal. The surface is inclined at 30° to horizontal, so its normal makes 60° with horizontal (or 30° with vertical). The angle between the force and the surface normal = 60° - 60° = 0°? Let me recalculate: Force at 60° from horizontal. Surface normal at 90°-30° = 60° from horizontal. Angle between force and normal = 60° - 60° = 0°. Component perpendicular to surface = 100·cos(0°) = 100 N. Alternatively: angle between force and surface = 60° - 30° = 30°. Component perpendicular = 100·sin(90°-30°) = 100·cos(30°) = 86.6 N. Rechecking: force at 60° to horizontal, surface at 30° to horizontal. Angle of force with surface = 60°-30° = 30°. Normal component = F·sin(30°) = 100×0.5 = 50 N.

Explanation:
For a coplanar (2D) force system, there are 3 independent equilibrium equations: (1) ΣFx = 0 (sum of horizontal forces = 0), (2) ΣFy = 0 (sum of vertical forces = 0), (3) ΣM = 0 (sum of moments about any point = 0). For a 3D (spatial) force system, there are 6 equilibrium equations: ΣFx=0, ΣFy=0, ΣFz=0, ΣMx=0, ΣMy=0, ΣMz=0.

Explanation:
Let angle between P and Q be θ. Original resultant: R² = P² + Q² + 2PQ·cos(θ). When Q is doubled (2Q), new resultant R' is perpendicular to P. For R' ⊥ P: P + 2Q·cos(θ) = 0 (horizontal component of R' = 0). So cos(θ) = -P/(2Q). Also, R² = P² + Q² + 2PQ·(-P/2Q) = P² + Q² - P² = Q². Therefore R = Q.

Explanation:
A general force system can be reduced to a single force (without a couple) when the resultant couple M = 0. If M ≠ 0 and R ≠ 0, the system reduces to a force and a couple (or a wrench in 3D). If R = 0 and M ≠ 0, it reduces to a pure couple. If both R = 0 and M = 0, the system is in equilibrium. So M = 0 is the condition for reduction to a single force.

Explanation:
When forces P and Q act at angle θ, the resultant R makes angle α with P. Resolving: R·cos(α) = P + Q·cos(θ) and R·sin(α) = Q·sin(θ). Dividing: tan(α) = Q·sin(θ)/(P + Q·cos(θ)). Therefore α = arctan[Q·sin(θ)/(P + Q·cos(θ))]. This is the standard formula for the direction of the resultant of two forces.

Explanation:
Forces along the plane: Applied force P = W·sin(θ) + μ·W·cos(θ) = 500·sin(30°) + 0.3×500·cos(30°) = 500×0.5 + 0.3×500×0.866 = 250 + 129.9 = 379.9 N. Normal reaction N = W·cos(θ) = 500×cos(30°) = 433 N. Friction force (opposing upward motion) = μN = 0.3×433 = 129.9 N. Total force = 250 + 129.9 = 379.9 N.

Explanation:
The second moment of area (moment of inertia) of a circle of diameter D about its centroidal axis: I = πD⁴/64 = πr⁴/4 (where r = D/2). About a diameter: I = πD⁴/64. Polar moment of inertia J = πD⁴/32 = 2I. The section modulus Z = I/y_max = (πD⁴/64)/(D/2) = πD³/32. These are fundamental formulas in mechanics of materials.

Explanation:
For a body in equilibrium under three non-parallel forces, the forces must be concurrent (meeting at a single point). This is because if they were not concurrent, there would be a net moment about the intersection of any two of them, violating equilibrium. This is a fundamental theorem: three non-parallel forces in equilibrium must be concurrent and coplanar.

Explanation:
A force F and a couple M (in the same plane) can always be reduced to a single force. The single force has the same magnitude and direction as F, but its line of action is shifted by a distance d = M/F from the original line of action. This shift moves the line of action so that the moment of the new force about the original point equals M. The couple is absorbed by shifting the force.

Explanation:
Weight W = 10 × 10 = 100 N. Component along plane (down) = W·sin(30°) = 100 × 0.5 = 50 N. Normal reaction N = W·cos(30°) = 100 × 0.866 = 86.6 N. Maximum static friction = μN = 0.4 × 86.6 = 34.64 N. Since the driving force (50 N) > maximum friction (34.64 N), the block slides. But if in equilibrium, friction = 50 N. Since max friction = 34.64 N < 50 N, block slides and friction = 34.64 N (kinetic). Answer: 34.64 N.