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MechYr2 Chapter 7 :: Applications of Forces

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1 MechYr2 Chapter 7 :: Applications of Forces
Last modified: 7th January 2019

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3 Overview 1:: Unknown forces for bodies in equilibrium.
There is nothing new in this chapter – it just brings together all the individual components we have learnt so far regarding forces: friction, components of forces, 𝐹=𝑚𝑎, inclined planes and connected particles, for different common types of problems. 1:: Unknown forces for bodies in equilibrium. 2:: Static problem involving weight, tension and pulleys “If the particle is in equilibrium, determine the magnitude of the force 𝑋.” 𝑋 20 𝑁 60° 30° 10𝑔 𝑁 3:: Objects in motion on inclined planes 4:: Connected particles requiring resolution of forces.

4 Finding unknown forces by resolving forces
4 𝑁 [Textbook] The diagram shows a particle in equilibrium under the forces shown. By resolving horizontally and vertically find the magnitudes of the forces 𝑃 and 𝑄. 𝑃 𝑁 45° 30° Just resolve separately in the horizontal and vertical directions, as before. 𝑄 𝑁 ? Fro Note: There are two ways of thinking about this. Either “overall force in horizontal direction is 0”, thus 𝑃 cos 30°−4 cos 45° =0, or “forces right = force left”, as I’ve used here. I think the latter is slightly less cumbersome. 𝑅 → : 𝑃 cos 30° =4 cos 45° 𝑅 ↑ : 𝑃 sin 30° +4 sin 45° =𝑄 𝑃= 4 cos 45 cos 30 = sf 𝑄= 4 cos 45 cos sin sin 45 =4.46 (3sf)

5 Unknown forces on inclined planes
[Textbook] The diagram shows a particle in equilibrium on an inclined plane under the forces shown. Find the magnitude of the force 𝑃 and the size of the angle 𝛼. ? 𝑅 ↗ : 𝑃 cos 𝛼 =8+5 sin 30° 𝑅 ↖ : 𝑃 sin 𝛼 +2=5 cos 30° When we have 𝑃 sin 𝛼 and 𝑃 cos 𝛼 , a good strategy is to divide them so we get just tan 𝛼 . You may have seen this in Year 2 trigonometry when expressing trig sums in the form 𝑅 sin⁡(𝑥+𝛼). 𝑃 sin 𝛼 =5 cos 30° −2 𝑃 cos 𝛼 =8+5 sin 30° ∴ tan 𝛼 = 5 cos 30° −2 8+5 sin 30° =0.222… 𝛼=12.5° (3sf) 𝑃 sin 12.5… =5 cos 30° −2 𝑃=10.8 (3sf) 2 𝑁 𝑃 𝑁 𝑃 sin 𝛼 ? 𝛼 𝑃 cos 𝛼 ? 8 𝑁 5 cos 30 ? 30° 5 𝑁 30° 5 sin 30 ? Recall that I recommend using dotted labelled arrow for the components of forces in diagrams.

6 Test Your Understanding
The diagram shows a particle in equilibrium on an inclined plane under the forces shown. Find the magnitude of the force 𝑄 and the size of the angle 𝛽. ? 𝑅 ↗ : 𝑄 cos 𝛽 =5+6 sin 30° 𝑅 ↖ : 𝑄 sin 𝛽 +6 cos 30° =14 𝑄 sin 𝛽 =14−6 cos 30° 𝑄 cos 𝛽 =5+6 sin 30° ∴ tan 𝛽 = 14−6 cos 30° 5+6 sin 30° 𝛽=47.7° (3sf) 𝑄 cos 47.7… =5+6 sin 30° 𝑄=11.9 𝑁 (3sf) 14 𝑁 𝑄 𝑁 𝛽 ? 5 𝑁 6 𝑁 30°

7 Exercise 7A Pearson Stats/Mechanics Year 2 Pages

8 Modelling with Statics
We can apply this to problems involving tension, weight and pulleys. [Textbook] A smooth bead 𝑌 is threaded on a light inextensible string. The ends of the string are attached to two fixed points, 𝑋 and 𝑌, on the same horizontal level. The bead is held in equilibrium by a horizontal force of magnitude 8 N acting parallel to 𝑍𝑋. The bead 𝑌 is vertically below 𝑋 and ∠𝑋𝑍𝑌=30° as shown in the diagram. Find the tension in the string and the weight of the bead. 𝑋 𝑍 30° 8 𝑁 𝑌 As the bead is smooth, the two parts of the string can be considered as a single piece of string, and therefore the tension is the same throughout. ? Force Diagram 𝑇 𝑁 𝑇 𝑁 ? 8 𝑁 30° 𝑅 → : 𝑇 cos 30° =8 ∴ 𝑇=9.24 (3sf) 𝑅 ↑ : 𝑊=𝑇+𝑇 sin 30° =13.9 (3sf) ? 𝑊 𝑁

9 Further Example ? ? Force Diagram ? ? ? a For 1kg mass: 𝑅 ↑ : 𝑇=𝑔
[Textbook] A mass of 3kg rests on the surface of a smooth plane which is inclined at an angle of 45° to the horizontal. The mass is attached to a cable which passes up the plane along the line of greatest slope and then passes over a smooth pulley at the top of the plane. The cable carries a mass of 1kg freely suspended at the other end. The masses are modelled as particles, and the cable as a light inextensible string. There is a force of 𝑃 N acting horizontally on the 3kg mass and the system is in equilibrium. Calculate (a) the magnitude of 𝑃 (b) the normal reaction between the mass and the plane (c) State how you have used the assumption that the pulley is smooth in your calculations. a For 1kg mass: 𝑅 ↑ : 𝑇=𝑔 For 3kg mass: 𝑅 ↗ : 𝑇+𝑃 cos 45° =3𝑔 sin 45° ∴𝑃= 3𝑔 sin 45° −𝑔 cos 45° =3𝑔−𝑔 2 =16 (2sf) 𝑅 ↖ : 𝑅=3𝑔 cos 45° +𝑃 sin 45° =6𝑔 −𝑔=32 (2sf) Pulley is smooth so tension in string will be same on both sides of the pulley. ? Force Diagram 𝑅 ? 𝑇 𝑃 𝑇 45° 𝑃 sin 45 3𝑔 cos 45 ? 1𝑔 𝑃 cos 45 3𝑔 45° 3𝑔 sin 45 b ? c ?

10 Test Your Understanding
Edexcel M1(Old) May 2013(R) Q2 The particle can’t move along the string, so we have two separate strings with separate tensions. Introduce suitable variables for the tensions of each, e.g. 𝑇 1 and 𝑇 2 . ?

11 Exercise 7B Pearson Stats/Mechanics Year 2 Pages

12 Friction ? Force Diagram ? Working ? Force Diagram ? Working a
Earlier in the module we saw that the frictional force 𝐹≤𝜇𝑅, where 𝐹=𝜇𝑅 if the object on the plane is moving. Were the object is not moving, we saw that the force of friction acts in a direction opposite to that which it would be moving if the frictional force wasn’t there. [Textbook] A box of mass 10kg rests in limiting equilibrium on a rough plane inclined at 20° above the horizontal. Find the coefficient of friction between the box and the plane. A horizontal force of magnitude 𝑃 N is applied to the box. Given that the box remains in equilibrium, (b) find the maximum possible value of 𝑃. a ? Force Diagram 𝑅 ? Working 𝑅 ↖ : 𝑅=10𝑔 cos 20° 𝑅 ↗ : 𝜇𝑅=10𝑔 sin 20° 𝜇= 10𝑔 sin 20° 10𝑔 cos 20° = tan 20° =0.36 (2𝑠𝑓) 𝜇𝑅 10𝑔 cos 20 10𝑔 20° 20° 10𝑔 sin 20 b ? Force Diagram 𝑅 ? Working 𝑅 ↖ : 𝑅=10𝑔 cos 20° +𝑃 sin 20° 𝑅 ↗ : 𝑃 cos 20° =10𝑔 sin 20° +𝜇𝑅 𝑃 cos 20° =10𝑔 sin 20° 𝑔 cos 20° +𝑃 sin 20° … 𝑃=82 𝑁 (2sf) Important: If we increase 𝑃 to its maximum, the particle is on the verge of moving UP the plane. Thus friction acts downwards to oppose. 𝑃 20° 𝑃 sin 20 10𝑔 cos 20 𝑃 cos 20 𝜇𝑅 10𝑔 20° 20° 10𝑔 sin 20

13 Test Your Understanding
Edexcel M1(Old) Jan 2006 Q5 ? a ? b ? c

14 Exercise 7C Pearson Stats/Mechanics Year 2 Pages

15 Static rigid bodies ? ? Diagram ? ?
Recall from the chapter on moments that for a stationary rigid body: The resultant force is 0. The resultant moment is 0. The problems are the same as in the moments chapter, except now we may need to consider frictional forces. [Textbook] A uniform rod 𝐴𝐵 of mass 40kg and length 10m rests with the end 𝐴 on rough horizontal ground. The rod rests against a smooth peg 𝐶 where 𝐴𝐶=8 m. The rod is in limiting equilibrium at an angle of 15° to the horizontal. Find: the magnitude of the reaction of 𝐶 the coefficient of friction between the rod and the ground. ? Diagram a Taking moments about A: 40𝑔×5 cos 15° =𝑁×8 𝑁=236.65… The reaction at 𝐶 has magnitude 240N (2sf). 𝑅 → : 𝜇𝑅=𝑁 sin 15° =61.25 N 𝑅 ↑ : 𝑅+𝑁 cos 15° =40𝑔 ∴𝑅=40𝑔−𝑁 cos 15° = ∴𝜇= =0.37 (2sf) 𝑁 ? 15° 2 m 𝐵 𝑅 𝐶 3 m ? 5 m b 40𝑔 This frictional force prevents the rod slides against the ground. ? 15° 𝜇𝑅 𝐴 5 cos 15° m

16 Further Example ? ? ? Diagram ? ? a b 𝑅 → :𝜇𝑅=𝑃 𝑅 ↑ : 𝑅=2𝑚𝑔+𝑚𝑔=3𝑚𝑔
[Textbook] A ladder 𝐴𝐵, of mass 𝑚 and length 3𝑎, has one end 𝐴 resting on rough horizontal ground. The other end 𝐵 rests against a smooth vertical wall. A load of mass 2𝑚 is fixed on the ladder at the point 𝐶, where 𝐴𝐶=𝑎. The ladder is modelled as a uniform rod in a vertical plane perpendicular to the wall and the load is modelled as a particle. The ladder rests in limiting equilibrium at an angle of 60° at an angle of 60° with the ground. Find the coefficient of friction between the ladder and the ground. State how you have used the assumption that the ladder is uniform in your calculations. ? Diagram a 𝑅 → :𝜇𝑅=𝑃 𝑅 ↑ : 𝑅=2𝑚𝑔+𝑚𝑔=3𝑚𝑔 Taking moments about 𝐵: 2𝑚𝑔×2𝑎 cos 60° + 𝑚𝑔×1.5𝑎 cos 60° + 𝜇𝑅×3𝑎 sin 60° =𝑅×3𝑎 cos 60° 2.75𝑚𝑔 𝜇𝑅=1.5𝑅 2.75𝑚𝑔 𝜇 3𝑚𝑔 =1.5 3𝑚𝑔 𝜇=4.5 𝜇=0.225 (3sf) You can assume its weight acts at its midpoint. ? 𝑃 𝐵 ? 1.5𝑎 ? 0.5𝑎 𝐶 𝑚𝑔 𝑅 𝑎 2𝑚𝑔 60° 𝐴 𝐹 As before, two different reaction forces so need different variables. b ?

17 Exercise 7D Pearson Stats/Mechanics Year 2 Pages

18 Particles moving on a rough plane
We’ve previously considered particles moving on smooth planes. And particles in equilibrium (and not moving) on rough planes. So let’s consider particles moving on rough planes! [Textbook] A particle is held at rest on a rough plane which is inclined to the horizontal at an angle 𝛼, where tan 𝛼 = The coefficient of friction between the particle and the plane is 0.5. The particle is released and slides down the plane. Find: the acceleration of the particle. the distance it slides in the first 2 seconds. ? Diagram If tan α= then sin α = 3 5 and cos α= 4 5 𝑅 ↖ : 𝑅=𝑚𝑔 cos 𝛼 𝑅= 4 5 𝑚𝑔 𝑅 ↙ : 𝑚𝑔 sin 𝛼 −𝜇𝑅=𝑚𝑎 𝑚𝑔− 2 5 𝑚𝑔=𝑚𝑎 𝑎=0.2𝑔=2.0 m s −2 𝑠=?,𝑢=0, 𝑣= ,𝑎=0.2𝑔, 𝑡=2 𝑠=𝑢𝑡+ 1 2 𝑎 𝑡 2 =3.9 (3sf) ? a 5 5 𝑎 3 3 𝑅 𝛼 𝜇𝑅 4 4 m𝑔 cos 𝛼 𝑚𝑔 𝛼 𝛼 𝑚𝑔 sin 𝛼 ? b

19 Test Your Understanding
Edexcel M1(Old) May 2013(R) Q5 ? ? ?

20 Exercise 7E Pearson Stats/Mechanics Year 2 Pages

21 Connected particles involving friction
We have already encountered problems involving connected particles in Mechanics Year 1. We just now throw friction into the mix. [Textbook] Two particles 𝑃 and 𝑄 of masses 5kg and 10kg respectively are connected by a light inextensible string. The string passes over a small smooth pulley which is fixed at the top of a rough inclined plane. 𝑃 rests on the inclined plane and 𝑄 hangs on the edge of the plane with the string vertical and taut. The plane is inclined to the horizontal at an angle 𝛼 where tan 𝛼 =0.75, as shown in the diagram. The coefficient of friction between 𝑃 and the plane is 0.2. The system is released from rest. Find the acceleration of the system. Find the tension in the string. 𝑃 5kg 𝑄 10kg 𝛼 sin 𝛼 = 3 5 , cos 𝛼 = 4 5 For 𝑃: 𝑅 ↖ : 𝑅=5𝑔 cos 𝛼 𝑅=5𝑔× 4 5 =4𝑔 𝑅 ↗ :𝑇−5𝑔 sin 𝛼 −0.2𝑅=5𝑎 𝑇−3.8𝑔=5𝑎 (1) At 𝑄: 𝑅 ↓ : 10𝑔−𝑇=10𝑎 (2) Adding (1) and (2): 6.2𝑔=15𝑎 → 𝑎= 31𝑔 75 =4.1 m s −2 Rearranging (1): 𝑇=5𝑎+3.8𝑔=57 N (2sf) a ? ? Diagram 5 5 𝑎 3 3 𝑅 𝑎 𝛼 𝑇 4 𝑃 5kg 𝑇 4 0.2𝑅 m𝑔 cos 𝛼 𝑄 10kg 𝑚𝑔 𝛼 10𝑔 𝛼 𝑚𝑔 sin 𝛼 b ?

22 Further Example [Textbook] One end of a light inextensible string is attached to a block 𝐴 of mass 2kg. The block 𝐴 is held at rest on a smooth fixed plane which is inclined to the horizontal at an angle of 30°. The string lies along the line of greatest slope of the plane and passes over a smooth light pulley which is fixed at the top of the plane. The other end of the string is attached to a block 𝐵 of mass 5kg. The system is released from rest. By modelling the blocks as particles and ignoring air resistance, (a)(i) show that the acceleration of block 𝐵 is 4 7 𝑔 (ii) find the tension in the string. (b) State how you have used the fact that the string is inextensible in your calculations. (c) Calculate the magnitude of the force exerted on the pulley by the string. 𝐴 2kg 𝐵 5kg 30° 𝑎 a ? For 𝐴: 𝑅 ↗ : 𝑇−2𝑔 sin 30° =2𝑎 𝑇=2𝑎+𝑔 (1) At 𝐵: 𝑅 ↓ : 5𝑔−𝑇=5𝑎 (2) Adding (1) and (2): 5𝑔=7𝑎+𝑔 → 𝑎= 4 7 𝑔 𝑇=2𝑎+𝑔= 8 7 𝑔+𝑔= 15 7 𝑔 String is inextensible so acceleration at 𝐴 and 𝐵 the same. 𝑅 𝑇 𝑎 𝐴 2kg 𝑇 m𝑔 cos 30 𝐵 5kg 𝑚𝑔 𝛼 5𝑔 30 𝑚𝑔 sin 30 b ? c ? Recall that tension acts away from object in direction of string, so two tensions acting on pulley. We want the resultant force. 𝐹𝑜𝑟𝑐𝑒= 𝑇 cos 30° 𝑇+𝑇 sin 30° = 𝑇 𝑀𝑎𝑔𝑛𝑖𝑡𝑢𝑑𝑒= 𝑇 = 𝑔× 12 = 𝑔 N We can consider left and down directions the positive ones to avoid negatives. 𝑇 is constant so can factor out. 30° 𝑇 𝑇

23 Test Your Understanding
? Edexcel M1(Old) May 2013(R) Q3

24 Exercise 7F Pearson Stats/Mechanics Year 2 Pages


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