Questions on statics of a particle

Question 1

State the condition for a particle to be in equilibrium, and how it is used in practice.

Accepted Answer

The resultant force on the particle is zero. In practice that means resolving in two perpendicular directions and setting each component sum to zero. B1 for the zero resultant, B1 for resolving in two perpendicular directions. Two directions give two equations, which is enough for two unknowns.

Question 2

A particle of weight 30 N rests on a smooth slope inclined at 25°, held by a string parallel to the slope. Find the tension and the normal reaction.

Accepted Answer

Resolving along the slope, T = 30 sin 25° = 12.7 N. Perpendicular to the slope, R = 30 cos 25° = 27.2 N. M1 for resolving along the slope, A1 for T = 12.7 N, M1 for resolving perpendicular to it, A1 for R = 27.2 N. Choosing the slope and its normal as the two directions keeps the unknowns apart, because each equation then contains only one of them. Resolving horizontally and vertically instead would give a pair of simultaneous equations for the same answer.

Question 3

Forces (3i + 2j) N, (−5i + 4j) N and F act on a particle which is in equilibrium. Find F, and its magnitude.

Accepted Answer

Equilibrium means the three forces sum to zero, so F = −[(3 − 5)i + (2 + 4)j] = −(−2i + 6j) = (2i − 6j) N. Its magnitude is √(2² + 6²) = √40 = 6.32 N (3 s.f.). M1 for setting the vector sum to zero, A1 for F, A1 for 6.32 N. The i components and the j components close separately: 3 − 5 + 2 = 0 and 2 + 4 − 6 = 0, the two scalar equations inside one vector statement.

Question 4

Three forces of magnitudes 5 N, 12 N and 13 N act on a particle which is in equilibrium. Explain why the 5 N and 12 N forces must be perpendicular to each other.

Accepted Answer

In equilibrium the 13 N force balances the resultant of the other two, so that resultant has magnitude 13 N. Since 5² + 12² = 25 + 144 = 169 = 13², the resultant of the 5 N and 12 N forces has magnitude 13 N exactly when they are at right angles, by Pythagoras; at any other angle the cosine rule gives a magnitude other than 13. So the two forces are perpendicular. B1 for the resultant of two forces balancing the third, B1 for the Pythagoras argument. The magnitude of the resultant of two forces depends on the angle between them, ranging from 7 N when they oppose to 17 N when they align.

Question 5

A 10 N weight hangs in equilibrium from two strings making 20° and 70° with the horizontal. Find both tensions.

Accepted Answer

Horizontally, T₁ cos 20° = T₂ cos 70°. Vertically, T₁ sin 20° + T₂ sin 70° = 10. Solving the pair gives T₁ = 3.42 N and T₂ = 9.40 N. The steeper string carries far more of the load, which the sketch should have led you to expect. The angles sum to 90°, so the strings are perpendicular, the weight is being split into two perpendicular components, and Pythagoras must reassemble it: 3.42² + 9.40² = 100 = 10². That is M1 M1 for the two resolving equations, dM1 for solving them and A1 A1 for the two tensions.

Question 6

A particle of weight 40 N is held at rest on a smooth 30° slope by a horizontal force P. Find P and the normal reaction.

Accepted Answer

The reaction is perpendicular to the slope, so it makes 30° with the vertical. Resolving horizontally, R sin 30° = P. Resolving vertically, R cos 30° = 40. Dividing the first by the second gives P = 40 tan 30° = 23.1 N, and then R = 40/cos 30° = 46.2 N. M1 for resolving horizontally, A1 for R sin 30° = P, M1 for resolving vertically, A1 for P = 23.1 N, A1 for R = 46.2 N. The reaction exceeds the weight, because pressing horizontally into a slope squeezes the surface harder than gravity alone does. Any answer with R below 40 N has resolved the wrong way round.

Question 7

A particle of mass 3 kg rests in limiting equilibrium on a rough slope inclined at 35° to the horizontal. Find the coefficient of friction between the particle and the slope.

Accepted Answer

Resolving perpendicular to the slope, R = 3g cos 35°. Resolving along it, friction acts up the slope and F = 3g sin 35°. Limiting equilibrium means F = μR, so μ = tan 35° = 0.700. M1 for resolving perpendicular to the slope, M1 for resolving along it, M1 for using F = μR in limiting equilibrium, A1 for μ = 0.700. The mass cancels, so any particle on the point of slipping on this slope gives the same answer: tilting a surface until an object just slides is a practical way of measuring μ, and the angle it slides at is independent of how heavy the object is.

Question 8

A particle of weight 50 N is held in equilibrium on a smooth slope inclined at 30° to the horizontal by a string inclined at 20° to the slope. Find the tension in the string and the normal reaction.

Accepted Answer

Keep the axes along and perpendicular to the slope, and split the tension across them. Along the slope, T cos 20° = 50 sin 30° = 25, so T = 26.6 N. Perpendicular to it, the string now helps hold the particle off the surface: R + T sin 20° = 50 cos 30°, so R = 43.3 − 9.1 = 34.2 N. M1 for resolving along the slope, A1 for T = 26.6 N, M1 for resolving perpendicular with the T sin 20° term included, A1 for R = 34.2 N, B1 for a diagram or statement showing the tension acting at 50° to the horizontal, away from the particle. Compare the parallel-string case: there T would be exactly 25 N. Angling the string upward costs tension, because only the cos 20° part holds the slope direction, but it relieves the surface, which is why R has dropped below 50 cos 30° = 43.3 N. Omitting T sin 20° and writing R = 43.3 N is the standard error.

Question 9

A block of mass 6 kg rests on a rough slope inclined at 15° to the horizontal, with coefficient of friction 0.3. Show that the block remains in equilibrium, and state the magnitude of the friction force acting on it.

Accepted Answer

Equilibrium along the slope demands a friction force of 6g sin 15° = 15.2 N acting up the slope. The most friction the surface can supply is μR = 0.3 × 6g cos 15° = 0.3 × 56.8 = 17.0 N. Since 15.2 N < 17.0 N, the demand is within the ceiling and the block stays put. The friction force is 15.2 N, not 17.0 N. M1 for resolving along the slope, A1 for the 15.2 N needed, M1 for resolving perpendicular and forming μR, A1 for the 17.0 N ceiling, B1 for the comparison and for stating F = 15.2 N. Friction is sized by what the other forces demand and capped at μR; only limiting equilibrium makes the cap an equality.

Question 10

A particle of weight 20 N hangs from a light string. A horizontal force H is applied to the particle so that it rests in equilibrium with the string making 40° with the vertical. Find the tension in the string and the value of H.

Accepted Answer

Resolve vertically: T cos 40° = 20, so T = 20/cos 40° = 26.1 N. Resolve horizontally: H = T sin 40° = 26.1 × 0.643 = 16.8 N, or directly H = 20 tan 40° = 16.8 N. M1 for resolving vertically, A1 for T = 26.1 N, M1 for resolving horizontally, A1 for H = 16.8 N. The angle is measured from the vertical, so cos 40° goes with the vertical component; a string at 40° to the horizontal would give T = 20/sin 40° instead. The tension exceeds the weight because the string holds the particle sideways as well as up.

Question 11

A particle of mass 5 kg rests on a rough horizontal floor, coefficient of friction 0.4. A force of 15 N is applied to the particle at 30° above the horizontal. Taking g = 9.8 m s⁻², show that the particle remains at rest, and find the friction force acting on it.

Accepted Answer

Vertically, R + 15 sin 30° = 5g, so R = 49 − 7.5 = 41.5 N, and the friction limit is 0.4 × 41.5 = 16.6 N. Horizontally, equilibrium needs friction equal to 15 cos 30° = 13.0 N. Since 13.0 < 16.6, the surface can supply it and the particle stays at rest, with friction 13.0 N acting against the pull. M1 for resolving vertically with the upward component of the 15 N, A1 for the limit 16.6 N, M1 for comparing the horizontal component with the limit, A1 for F = 13.0 N. Friction is 13.0 N and not 16.6 N: in equilibrium it matches the demand, and the limit is reached only on the point of slipping.

Question 12

A particle of mass 4 kg rests on a rough horizontal plane with coefficient of friction 0.25. A force P is applied at 20° above the horizontal. Find the least value of P that will move the particle.

Accepted Answer

On the point of moving the particle is in limiting equilibrium, which is rest at the very moment of slipping. That is what adds the third equation F = μR to the two resolving equations, and it is the only circumstance in which friction may be written at its maximum. Resolving vertically, the angled force lifts as well as pulls: R + P sin 20° = 4g, so R = 39.2 − P sin 20°. Resolving horizontally at the limit, P cos 20° = μR = 0.25(39.2 − P sin 20°). Expanding, P cos 20° + 0.25P sin 20° = 9.8, so P(0.9397 + 0.0855) = 9.8 and P = 9.8/1.0252 = 9.56 N. The reaction is then 35.9 N, well below the weight of 39.2 N. B1 for stating limiting equilibrium with F = μR, M1 for resolving vertically, A1 for R = 39.2 − P sin 20°, M1 for resolving horizontally at the limit, A1 for the equation in P, dM1 for solving it, A1 for P = 9.56 N. Pulling at an angle lightens the load on the surface, lowers the friction ceiling, and so needs less force than the 9.8 N a purely horizontal pull would require.

Question 13

A block of mass 8 kg rests on a rough slope inclined at 30° to the horizontal, with coefficient of friction 0.2, held by a force P acting up the slope. Find the least and greatest values of P for which the block remains in equilibrium.

Accepted Answer

Perpendicular to the slope, R = 8g cos 30° = 67.9 N, so the friction ceiling is μR = 13.6 N. Down the slope the weight component is 8g sin 30° = 39.2 N. Friction reverses direction between the two ends of the range, so draw its arrow both ways before writing either equation. About to slip down, friction acts up the slope at full strength: P + 13.6 = 39.2, so the least P is 25.6 N. About to slip up, friction turns to act down the slope: P = 39.2 + 13.6 = 52.8 N is the greatest. Equilibrium therefore holds for 25.6 ≤ P ≤ 52.8 N. M1 for resolving perpendicular, A1 for μR = 13.6 N, B1 for friction acting in opposite directions in the two limiting cases, M1 for the slipping-down equation, A1 for 25.6 N, M1 for the slipping-up equation, A1 for 52.8 N. The lower end is positive because tan 30° = 0.58 exceeds μ = 0.2, so the block cannot rest unaided: without P it would slide, which is why a holding force is needed at all.

Question 14

A particle of weight W hangs in equilibrium from two light strings attached to a fixed horizontal ceiling, the strings making angles p and q with the horizontal. By resolving horizontally and vertically, show that the tension in the string at angle p is W cos q / sin(p + q). Deduce the two tensions in the case p + q = 90°. Use the general result to find both tensions when W = 12 N, p = 40° and q = 60°, and verify that they support the weight.

Accepted Answer

Let T₁ be the tension at angle p and T₂ at angle q. Horizontally: T₁ cos p = T₂ cos q. Vertically: T₁ sin p + T₂ sin q = W. From the first, T₂ = T₁ cos p / cos q. Substituting: T₁(sin p + cos p sin q / cos q) = W, so T₁(sin p cos q + cos p sin q) = W cos q, and sin p cos q + cos p sin q = sin(p + q), giving T₁ = W cos q / sin(p + q). By the same working T₂ = W cos p / sin(p + q). When p + q = 90°, sin(p + q) = 1, so T₁ = W cos q and T₂ = W cos p. With W = 12, p = 40° and q = 60°, p + q = 100°, so the general form is needed: sin 100° = 0.985, giving T₁ = 12 cos 60°/sin 100° = 6/0.985 = 6.09 N and T₂ = 12 cos 40°/sin 100° = 9.19/0.985 = 9.33 N. Check vertically: 6.09 sin 40° + 9.33 sin 60° = 3.92 + 8.08 = 12.0, the weight; and horizontally 6.09 cos 40° = 4.67 = 9.33 cos 60°. M1 for the horizontal equation, M1 for the vertical equation, M1 for eliminating T₂, A1 for the compound-angle step to the required form, A1 for the perpendicular case, A1 for 6.09 N and 9.33 N with the check. The string at the larger angle to the horizontal has the larger cosine of the other angle and so carries more of the weight; the formula makes that general rather than numerical. These strings are not perpendicular, so the tensions do not obey Pythagoras: 6.09² + 9.33² = 124, not 144, and the shortcut that works for a perpendicular pair has no place once p + q is not 90°.

Question 15

A particle of weight 50 N rests in equilibrium on a smooth plane inclined at 30° to the horizontal, held by a light string making an angle k above the line of greatest slope, where 0° ≤ k < 90°. Explain why the tension is least when k = 0, and find the value of k for which the particle is on the point of leaving the surface of the plane. State the tension in that case.

Accepted Answer

Along the slope: T cos k = 50 sin 30° = 25, so T = 25/cos k. Since cos k is largest, equal to 1, when k = 0, the tension is least, 25 N, when the string is parallel to the slope, and it increases as the string is raised. Perpendicular to the slope: R + T sin k = 50 cos 30° = 43.3. The particle is on the point of leaving the surface when R = 0, so T sin k = 43.3 while T cos k = 25: dividing, tan k = 43.3/25 = √3 and k = 60°. The string then makes 30° + 60° = 90° with the horizontal, so it is vertical, and T = 25/cos 60° = 50 N, the whole weight. M1 for the along-slope equation, A1 for the least tension at k = 0 with the reason, M1 for the perpendicular equation with R = 0, A1 for k = 60°, B1 for the interpretation that the string is vertical and carries the whole weight. Beyond k = 60° the string would pull the particle off the plane, so the model of a particle resting on the surface holds only for k ≤ 60°.