Chapter 2 2.1 (i) y T(y) L y M T(y+y)+ yg L (ii) Since the element of length y is in equilibrium, we have M T ( y) T ( y y) yg L Using Taylor series expansion for T(y+y), which gives dT 1 d 2T T ( y y) T ( y) y 2 (y) 2 dy 2 dy And taking limit y 0 leads to the differential equation for T(y). The equation is dT M g dy l Solution of this equation is M T ( y) gy C L where C is the integration constant. This is determined by using the fact that at the lose end (y=L) of the rope, the tension is zero. This gives M ( L y) C Mg and T ( y) g L 2.2 Torque r F ˆ j It is given that r 2i ˆ and the force has magnitude 50N and acts in the direction ˆ of vector 3i 2 ˆ . Thus the force is 50 times the unit vector in the direction of the j given vector. This gives ˆ 3i 2 ˆ j 13 F 50 ˆ j With this the torque is 2i ˆ 50 ˆ 3i 2 ˆ j 50 ˆ k 13 13 2.3 (a) 100N 100N (b) The centre of the rod is at (3, 2) ˆ The right end is at position r 8i 2 ˆ and the force at this end is j 3 F 100 ˆ 1 ˆ 2 i 2 j ˆ The left end is at position r 2i 2 ˆ and the force at this end is j 3 F 100 ˆ 1 i ˆ j 2 2 Torque with respect to the origin = 8iˆ 2 ˆj 100 3ˆ 1 i ˆ 2i 2 ˆ 100 3 i 1 ˆ 10i 100 3 i 1 ˆ j ˆ j 2 ˆ j ˆ 2 ˆ j 2 2 2 2 500k ˆ Torque with respect to the centre of the rod = 3 3 3 8i 100 ˆ ˆ 1 ˆ ˆ ˆ 1 ˆ ˆ ˆ 1 ˆ 2 i 2 j 2i 100 2 i 2 j 10i 100 2 i 2 j 500k ˆ Torque with respect to the left end of the rod = 3 10 i 100 ˆ ˆ 1 ˆ 2 i 2 j 500 k ˆ Torque with respect two the right end of the rod = 3 10 i 100 ˆ ˆ 1 ˆ 2 i2 j 500 k ˆ (c) Torques about all the points are equal because the net force on the rod is zero. 2.4 (a) TA TB 3m A B 1m 0.5m 30N 70N 120N (b) F y 0 gives TA TB 220 (i) Net torque about A is zero, which gives 3TB 1 70 1.5 30 2.5 120 (ii) 415 Equation (ii) gives TB 138 N This substituted in equation (i) gives TA 82 N 2.5 Component of force in the plane perpendicular to the axis is F cos at a distance of R from the axis. Therefore the torque about the axis is RF cos 2.6 Free-body diagram of the block N1 N2 W N1, N2 and W are three forces in a plane. Thus they must pass through one common point for equilibrium. So the equilibrium conditions are only the force conditions. F horizontal 0 gives N1 sin N 2 cos F vertical 0 gives N1 cos N1 sin W Solution of these two equations is N1 W cos and N 2 W sin 2.7 Free-body diagram of the plank N 2m Ry 1m 100N 0.2m Rx Free-body diagram of the block N F Nground If the rod makes angle with the horizontal then sin 0.2 and cos 0.98 (a) To get the horizontal force F, we first calculate the normal force N on the rod. To do so, we calculate the total torque about the hinged end of the plank 1 N 3 100 cos and equate it to zero. This gives N 3 100 cos 3 98 294N Now we balance the horizontal forces F horizontal 0 on the block to get F 294 sin 0.2 294 59 N (b) Force balance on the plank F horizontal 0 Rx N sin 59N F vertical 0 Ry N cos 100 This gives R y 100 288 188 N Minus sign in front implies that the direction is opposite to that shown in the free-body diagram above. 2.8 Free-body diagram of the rod N2 N1 F W Balancing the vertical forces gives N1 = W = 50N Balancing the horizontal forces gives N2 = F Balancing the torque about the centre of gravity gives F 8 50 0.5 leading to 25 F 8.8 N 8 2.9 Free body diagram of the painting T N1 N2 N1+N2 Fx W W Force balance equations give Fx T and N1 N 2 W N1 and N2 are equal because the component of torque perpendicular to the wall must vanish. This gives N1=N2=25N Balancing the component of torque parallel to the wall taken about the lower end of the painting gives 20 3 T 10 50 giving 25 T 14 .4 N 3 2.10 We first calculate the forces at the ends of the rod. These forces are applied by the supports. After finding the forces on the rod, we then calculate the forces and the torques applied by the wall on the supports. Free body diagram of the rod N1 N2 140cm 60cm 35N Free body diagrams of the left and the right supports F1 F2 1 2 5cm 5cm N1 N2 The forces on the rod satisfy Fy 0 which gives N1 N 2 35 Taking torque about the left end and using 0 gives 140 N 2 60 35 N 2 15 N This gives N1 20 N Now balancing vertical forces and the torque on the supports gives For the left support F1=20N and 1 0.05 20 1Nm For the right support F2=15N and 2 0.05 15 0.75 Nm 2.11 Ry 60cm Rx 40cm N 40N To find the force applied by the plastic block, we balance torque about the upper left corner. This leads to 40 N 30 40 N 30N Balancing the vertical forces gives Ry = 40N Balancing the horizontal forces gives Rx = 30N Negative sign means that the direction of Rx is opposite to that assumed in the free-body diagram above. Free-body diagram of the pole 30N 40cm 40N 30N N Balancing the vertical forces on the pole gives N = 40N There is no net horizontal force and the two horizontal forces give a couple = 300.4 = 12Nm Balancing the torques on the pole about the ground gives = 300.4 = 12Nm 2.12 Free-body diagram of the table 90cm ˆ j Rx Nx Ny Ry iˆ 20N To find Ny, we balance the torque on the table about its left hand edge to get 90 Ny 45 20 Ny 10N By balancing the vertical forces, we get Ry 10N . The negative sign tells us that the force is direction opposite to that shown above. Free-body diagram of one of the rods Ry 2 30 Rx 2 Sx Sy Free body diagram of the entire system 90cm 30 Nx Ny 20N 2Sy 2Sx To get Nx, we balance the component of the torque coming out of the paper on the entire system about the lower hinge. This gives 30 3 Nx 45 20 Nx 10 3N The negative sign again tells us that the direction of the force is opposite to that shown. Balancing the horizontal component of the force on the table then gives Rx 10 3N Rx ˆ Ry ˆ Note: The net force on each rod on its upper end is i ˆ j 5 3i 5 ˆ which is along j 2 2 the rod as it must be for the equilibrium of a rod held at its ends. Balancing the horizontal and vertical components of forces on each rod gives Rx Sx 5 3N and Sy 5N 2 Thus the net force on each rod is 10N compressive. 2.13 Free body diagrams of the two side portions and the portion AC over the pulley: N TA TC TA TC RM g L F L2 M g L L1 M g L Tension TA and TC at both ends of the portion over the pulley is the same because the torque L1 about the centre must vanish. This gives TA TC Mg L Free body diagrams of the portion AB and BC Neffy Neffy Neffx Neffx TB TB RM RM g g 2L 2L TA TC Notice that Torque of the normal reaction about the centre of the cylinder vanishes because for each small portion of the rope over the cylinder, the normal reaction is radial. Thus T A (or TC) and TB cannot be equal because they together provide a torque to balance the torque due to the weight of the rope. Balancing the torque about the centre on AB gives R RM R M R TA g R TB TB L1 g 2 2L 2 2 L Thus if the net force by the cylinder on the rope is Neff at an angle from the horizontal then by force balance R M R M N eff sin L1 g N eff cos L1 g 2 L 2 2 L Note that Neff acts at a point different from the centre of BC because on different infinitesimal portions it is different. 2.14 The support does not apply any torque about the x-axis. All other components and torques are balanced by the support. 2.15 When forces are applied at two points of the rod, force balance demands that the force be equal and opposite. However two such forces acting at two different points will give rise to a couple moment. The couple moment is zero only if the forces point along the rod (see figure below) Couple moment non-zero Couple moment zero 2.16 Let cables OA and OC make angle 1 and OB and OD angle 2 with the vertical. Then balancing the vertical forces gives 2T (sin 1 sin 2 ) 45000 The sine of the angles is easily calculated to be 1 1 2 sin 1 sin 2 2 11 4 5 This gives T=14050N 2.17 The torque direction is given by the direction of cross product n F , which is ˆ ˆ ˆ perpendicular to n . This implies there is no component of the torque in the direction of n . ˆ ˆ ˆ 2.18 The net force on the plate is 50 i 50 ˆ 70 i 120 i 50 ˆ j j Therefore the force that must be applied to the plate to keep it in equilibrium is ˆ F 120 i 50 ˆ . Since there are only three forces acting on the body, they must all pass j through the same point so that their net torque is zero. This is shown in figure below. A B O D C 50 ˆ The force F 120 i 50 ˆ is at an angle tan 1 j 22 .6 from the line DC. Thus it does 120 not pass through O and intersects with side AD and diagonal BD of the square. Therefore: (i) It is not possible to keep the square in equilibrium by applying the third force at O. (ii) It is possible to keep the square in equilibrium by applying the third force at a point on BD. Equation of BD with O as origin is y x a 5 3a Equation of line along which the third force acts is y x 2 12 2 3 Solving the two equations gives y x a 34 This gives the distance of point O = 0.125a 1 3 And from B = 2 a 0.58 a 2 34 (iii) It is clear that for equilibrium, the force can be applied only on AD and BC.