L 11 Force 3.doc

8Force 3Mechanical SystemsYou will recall the general advice given on solving force problems. Decide which object needs to be considered and draw a free body diagram for that object. That is,1. Draw a simple sketch of the object. (No need for artwork. If only translational motion is being considered the object can even be shown as a particle see Centre of Mass.)2. Show all the forces applied to the object by other bodies. Find the resultant force on the body or an expression for the resultant force on the body. This can be done using the triangle/polygon law for addition of vectors or by resolution. Note that if the direction of the acceleration is known the resultant force must be the sum of the components of the forces in that direction. Apply the equation F = ma.We will now see how this strategy is applied to some basic mechanical systems.Example 1 Simple Pulley(a)Assuming the string and pulley are ideal, calculate(i)the acceleration of each mass(ii) the tension in the string.(b)Calculate the tension in the string supporting the pulley and explain why it is less than the weight of 20.0 kg.m1m2m1 = 11.0 kgm2 = 9.0 kg(a) (i)m2m111g T = 11a(1)T -9g = 9a(2)(1) + (2) 2g = 20aa = 0.98 m s-2For m1 acceleration is 0.98 m s-2 upward.For m2 acceleration is 0.98 m s-2downward.(ii)From (1) T = 9(g + a) 97 N (b)Tension in string supporting pulley = 2 97 N = 194 N. This is less than the weight of 20 kg (196 N) because the centre of mass of the system is accelerating down. T (N)T (N)m1g (N) m2g (N) Example 2Bodies on an Inclined PlaneThe string, which is attached at X, passes along smooth holes through A, B and C over the smooth light pulley to D which hangs freely. (A, B and C are supported like beads on the string.)mA = 4.00 kg, mB = 5.00 kg, mC = 6.00 kg and mD = 8.00 kg. All surfaces in contact are smooth. Calculate(a) the acceleration of each body,(b) the tension in the string,(c) the force between A and B,(d) the force between B and C. 30.0oABCDX(a)Treating A, B and C as one body we have30o15g NDT8g NandT 15g sin 30o = 15 a(i)8g T = 8 a(ii)(i) + (ii)8g 15g sin 30o = 23 a a = 0.21 m s-2A, B and C accelerate up plane at 0.21 m s-2. D accelerates down at 0.21 m s-2. (b)From (ii)T = 8(g a) 76.7 N(c)Consider A.30o4g NT fBA 4g sin 30o = 4afBA = T 4g sin 30o 4a fBA = 56.2 NfBC 6g sin 30o = 6 0.21fBC = 6g sin 30o + 6 0.21 = 30.7 N(d)Consider C.6g NExample 3Bodies in Vertically Accelerated Frames of Reference (“lift problems”)A woman has a mass of 48 kg. She stands on a set of ordinary bathroom scales(using springs but calibrated in kg) on the floor of a lift. What should the scales read if:(a)the lift is stationary;(b)the lift is moving upwards at a steady 2.5 m s-1;(c)the lift accelerates upward at 2.0 m s-2;(d)the lift accelerates downward at 2.0 m s-2;(e) the lift cable breaks and the emergency brakes are not yet activated?(a)48 kg(b)48 kg(c)F mg = maF = m(a + g) = 48 11.8 NReading is 48 11.8 / 9.8 = 57.8 kg(d)mg F = maF = m(g a) = 48 7.8 NReading is 48 7.8/9.8 = 38.2 kg(e)mg F = mg so F is zero.Scales read zero.Examples1. A 4.00 kg mass is suspended below a 6.00 kg mass by a light string. The 6.00 kg mass is supported by another light string tied to the top. At a certain instant the tension in the upper string is 90.0 N.(a)Draw a diagram showing all forces acting on the 6.00 kg mass at this instant.(b)Find the acceleration of the 6.00 kg mass at this instant.(c) Find the tension in the lower string at this instant.2. An 8.00 kg mass is suspended from the ceiling by a light string. A 2.00 kg mass is hung from the bottom of the 8.00 kg mass by a long light spring.(a)What is the tension in the spring?(b)What is the tension in the string? The string is now cut.(c)For each mass, draw a diagram showing all the forces acting on that mass.(d) Calculate the initial acceleration of each mass.(e) Explain whether or not these accelerations will be constant.3.The blocks Q and R are connected by an ideal string that passes over an ideal pulley as shown in the diagram below.The mass of Q is 12 kg and the mass of R is 4.0 kg. A force F1 is applied to Q. This prevents the block from moving.(a)Calculate the magnitude of F1.(b)Calculate the tension in the string.F1 is removed and a force of F2 is applied to R and the blocks again are stationary.(c)What is now the tension in the string?(d)Calculate the force F2.F2 is removed and the system accelerates.(f) Calculate the tension in the string.4.Two bodies of equal masses, m kg, are joined by a light inextensible cord passing over a smooth light pulley. Find what fraction of the mass must be taken from one and added to the other so that each will travel 1.00 m in 2.00 s, when released from rest.5. This problem relates to the diagram below. The pulleys have frictionless bearings and the cords are light and inextensible.All surfaces are smooth.(a)Calculate the acceleration of the system.(b)Calculate the tension in the cord between the 10 kg and the 13 kg masses.(c)Calculate the tension in the cord between the 5 kg and 13 kg masses.6.Four masses m1, m2 m3 and m4 are arranged as shown in the diagram below. The strings S1, S2, S3, S4 and S5 are ideal and the system is in equilibrium. The weight of m1 is 1200 N.(a)Calculate the weights of m2, m3 and m4.(b)Calculate the tension in strings S3 and S5.1.(b) 0.8 m s-2 down(c) 36 N2.(a) 19.6 N(b) 98 N (d) 8 kg : 12.25 m s-2 down 2 kg : zero (e) They will not. Spring contracts, tension decreases.3.(a) 78 N up (b) 39 N (c) 118 N (d) 78 N down (e) 59 N4.5.1 %5.(a) 1.8 m s-2 to left(b) 67 N(c) 43.6 N6. (a) 919 N, 2 120 N, 2 255 N.(b) 771 N, 2 255 N.Example 4A block of mass 6.00 kg rests on a rough plane inclined at 44o to the horizontal. This block is held at rest by a light flexible string