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I need help with exercise and practice it.

PRACTICE IT Use the worked example above to help you solve this problem. m 56.3 kg sits on the left end of a seesaw A woman of mass -a plank of length L 4.68 m, pivoted in the middle as shown in the figure (a) First compute the torques on the seesaw about an axis that passes through the pivot point. Where should a man of mass M = 74.9 kg sit if the system (seesaw plus man and woman) is to be balanced? (b) Find the normal force exerted by the pivot if the plank has a mass of mpl = 12.6 kg N (c) Repeat part (a), but this time compute the torques about an axis through the left end of the plank. EXERCISE HINTS: GETTING STARTED IM STUCK! Suppose a 34.1-kg child sits 1.22 m to the left of center on the the problem you just solved in the PRACTICE same seesaw as IT section. A second child sits at the end on the opposite side, and the system is balanced. (a) Find the mass of the second child. kg (b) Find the normal force acting at the pivot point. Need Help? Read It Submit Answer Save Progress Practice Another Version fo and te un of heo e s acting an te ae PROBLEM A woman of mass m55.0 kg sits on the left end of a seesaw-a plank of length L-4.00 m, pivoted in the middle as in the figure. (a) First compute the torques on the seesaw about an axis that passes through the pivot point. Where should a man of mass M = 75.0 kg sit if the system (seesaw plus man and woman) is to be balanced? (b) Find the normal force exerted by the pivot if the plank has a mass of mo 12.0 kg. (c) Repeat part (a), but this time compute the torques about an axis through the left end of the plank STRATEGY In part (a), apply the second condition of equilibrium, Er-0, computing toorques around the pivot point The mass of the plank forming the seesaw is distributed evenly on either side of the pivot point, so the torque exerted by gravity on the plank, tgravity can be computed as if all the plank's mass is concentrated at the pivot point. Then rrty is zero, as is the torque exerted by the pivot, because their lever arms are zero. In part (b) the first condition of equilibrium, 0, must be applied. Part (c) is a repeat of part (a) showing that choice of a different axis yields the same answer. sOLUTION (A) Where should the man sit to balance the seesaw? Apply the secene candtion ot u to k by ng sum of the torues eai o e The nt tors are Lat 0+-May my)0 presen the mas dance fram the todarce the pivot Stve ne equts for nevaat m s.a . 750 k it. (B) Find the normal force exerted by the pivot on the seesaw m-A Apply the ist condatiun of wquilbrium to the plank soing he resuting for the urenown nema aIN+m (C) Repeat part (a), choosing a new axis through the left end of the plank. Campute the torous uning this a, and set heir oum equal to zers. Now the v and grety n p resu inangers ues ML2 mg)-m )- Subtute at anewn quantt -(12.0kg XaDm n0 -147 1Nm-175 N-3 (2.00mn e Saba besulertution the nrmad e 146 m

Answer 1

Srution M56.3 kg, .68 M= 74.9 ka Ca) 12.6kg Mp Mg Torues abont pivt 0 a mpg aib thnong pivol dve to wowan. (мa) (n) due to Man, Net tre i zeno. as System balan le (56.3) (.68) mL M5-mg - 2 M () (74.9) 58 m 76 m

FN (5) mg Mg (9 81) (7p.9+563 2.6) FNmg Mpg MS + ( Ais tknh lafte end of Plantk pFN Mg mpg mg 0 2 Tne dae to Mg 2end as mg ats f notntion fnengh (FA)( (M)) (M+m+ mp9 ) () ((+) ML M L 2M (56.3) (4.68) .76 m (274.9)

34.1 kg Mass of decon3 child = 22 m M (a) Sstem is balanced. Met rque about pivat is pRank zzens Torgne due t mas 2MCA (2) (3.1)1.22) 17.8 kg 7.7786 r. 68 () Noml fore at pivt +m's FNi FN (3.1 6) (.1) 2.6 17.778 O.633 x ID 632.535 N . FN