PHYS 210 Problems Chapter 5, Fall 2015

Chapter V Class problems Solution 12. From the slope of the graph we find a, = 3.0 m/s". Applying Newton’s second law to the x axis (and taking 6 to be the angle between F; and F;), we have F,+F,cos0= ma, => 0@0=56°. *P5.21 (a) Isolate either mass f ; T+mg=ma=0 > \T|= |mg| C) The scale reads the tension T, so T =mg=5.00 kg(9.80 m/s?)=]49.0N | oon (c) Isolate the pulley |F T,+27,=0 Ar T, = 2|T,|= 2mg =| 98.0 N NTL hin Gt (d) \F=i+T+mg=0 FIG. P5.24(c} Take the component along the incline 4 = neem N ao 0=30.0° or 0+T—-—mgsin30.0°= 0 bas 5.00(9.80 ‘ T = mesin30.0° = = a 490N , FIG. P5.21(d) =| 24.5N 55. The free-body diagrams for the two blocks in (a) are shown below. F is the applied force and Fee, is the force exerted by block | on block 2. We note that F is applied directly to block 1 and that block 2 exerts a force F. -F_,, on block | (taking Newton’s third law into account). onl — ond Newton’s second law for block | is F—F, 2onl =m, where a is the acceleration. The second law for block 2 is F,,,, =m,a. Since the blocks move together they have the same acceleration and the same symbol is used in both equations. (a) From the second equation we obtain the expression a = F,,,, /m,, which we substitute into the first equation to get F—F,,,, =m,F,,,,/m,. Since F,,,, =F,,,. (same magnitude for the third-law force pair), we obtain _ 1.2kg m+m, 2.3kg+1.2kg iI P| TT o (3.2 N)=1.1N. (b) If F is applied to block 2 instead of block 1 (and in the opposite direction), the free- body diagrams would look like the following: Fy, Fy, m, mM, =, bi — F, on2 F, on | F mg mg The corresponding force of contact between the blocks would be 2 i. =f.=—a—F 2355 (3.2 N)=2.1N. ton? mm, +m, 2.3 kg + 1.2kg (c) We note that the acceleration of the blocks is the same in the two cases. In part (a), the force F,,_, is the only horizontal force on the block of mass m, and in part (b) Fy, is the only horizontal force on the block with m, > m2. Since F,,.,=m,a in part (a) and Fj,,, = ma in part (b), then for the accelerations to be the same, Fy, > F,,,2- that is, the force between the blocks must be larger in part (b). Note: This problem demonstrates that while being accelerated together under an external force, the force between the two blocks is greater if the smaller mass is pushing against the bigger one. In the special case where m, =m, =m, Fj,,, = Fy,,, =F /2- 67. First we analyze the entire system with “clockwise” motion considered positive (that is, downward is positive for block C, rightward is positive for block B, and upward is positive for block 4): meg—m4g= Ma (where M= mass of the system = 24.0 kg). This yields an acceleration of a=g(mc — my)(M= 1.63 m/s’. Next we analyze the forces just on block C: mcg — T =mca. Thus the tension is T= mcg(2my+ mgVM=81.7N.