Phys 201A. Homework 6 Solutions. F A and F r. B. According to Newton s second law, ( ) ( )2. j = ( 6.0 m / s 2 )ˆ i ( 10.4m / s 2 )ˆ j.

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1 7. We denote the two foces F A + F B = ma,sof B = ma F A. (a) In unit vecto notation F A = ( 20.0 N)ˆ i and Theefoe, Phys 201A Homewok 6 Solutions F A and F B. Accoding to Newton s second law, a = [ ( 12 m / s 2 )sin( 30 )]ˆ i [( 12 m / s 2 )cos( 30 )]ˆ j = ( 6.0 m / s 2 )ˆ i ( 10.4m / s 2 )ˆ j. F B = ( 2.0 kg) 6.0m ( / s 2 )ˆ i + ( 2.0 kg) ( 10.4m / s 2 )ˆ j ( 20.0 N)ˆ i = 32 ( N)ˆ i + ( 21 N)ˆ j. (b) The magnitude of F B is ( 32 N) 2 + ( 21 N) 2 = 38.3 N. (c) The angle that F B makes with the positive x axis is found fom tan θ = F B y /F B x = ( 21 N)/( 32 N) = Consequently, the angle is eithe 33 o = 213. Since both the x and y components ae negative, the coect esult is In all thee cases the scale is not acceleating, which means that the two cods exet foces of equal magnitude on it. The scale eads the magnitude of eithe of these foces. In each case the tension foce of the cod attached to the salami must be the same in magnitude as the weight of the salami because the salami is not acceleating. Thus the scale eading is mg, whee m is the mass of the salami. Its value is (11.0 kg) (9.8 m/s 2 ) = 108 N. 17. The fee-body diagam fo the puck is shown below. N is the nomal foce of the ice on the puck, f kin is the kinetic foce of fiction (in the x diection), and mg is the foce of gavity. (a) The magnitudes of hoizontal component of Newton s second law gives f kin = ma, and constant acceleation kinematics (Table 2-1) can be used to find the acceleation whee Eq is solved fo ( t 2 t 1 ) and substituted into Eq to give v 22 = v a( x 2 x 1 ). Since the final velocity is zeo, this leads to v 12 = 2a x 2 x 1 substituted into the Newton s law equation to obtain mv 1 2 f kin = 2( x 2 x 1 ) kg = 215m = 0.13 N. ( ) 6.0m/s ( ) ( )2 ( ) a = v 12 2 x 2 x 1 ( ). This is

2 f kin mg (b) The magnitudes of the vetical components of Newton s second law gives N mg = 0, so N = mg which implies (using Eq. 6-10) that f kin = μ kin mg. We solve fo the coefficient: 0.13 N kg μ kin = f kin = ( ) ( 9.8 m / s 2 ) = (a) The fee-body diagam fo the block is shown below. F is the applied foce, N is the nomal foce of the wall on the block, f is the foce of fiction, and mg is the foce of gavity. To detemine if the block falls, we find the magnitude f stat of the static foce of fiction equied to hold it without acceleating ( μ stat N ) and also find the nomal foce of the wall on the block. We compae f and μ stat N. If f < μ stat N, the block does not slide on the wall but if f > μ stat N, the block does slide. The magnitude of the hoizontal component of Newton s second law is F N = 0, so N = F = 12 N and μ stat N = (0.60)(12 N) = 7.2 N. The vetical component is f mg = 0 N, so f = mg = 5.0 N. Since f < μ stat N the block does not slide. mg (b) Since the block does not move f = 5.0 N and N = 12 N. The foce of the wall on the block is F w b = N ˆ i + f ˆ j = ( 12 N ) ˆ i + ( 5.0 N) ˆ j

3 whee the axes ae as shown in Fig of the text. 23. (a) The fee-body diagam fo the cate is shown below. T is the tension foce of the ope on the cate, N is the nomal foce of the floo on the cate, mg is magnitude of the foce of gavity, and f is the foce of fiction. We take the +x diection to be hoizontal to the ight and the +y diection to be up. We assume the cate is motionless. The magnitudes of the x-components of Newton s second law leads to T cosθ f = 0 and the magnitude of the y-components becomes T sinθ + N mg = 0, whee θ = 15 is the angle between the ope and the hoizontal. mg The fist equation gives f = T cosθ and the second gives N = mg T sinθ. If the cate is to emain at est, f must be less than μ stat N, o T cosθ < μ stat ( mg T sinθ). When the tension foce is sufficient to just stat the cate moving, we must have T cosθ = μ stat ( mg T sinθ). We solve fo the magnitude of the tension: μ T = stat mg cos θ + μ stat sin θ ( = 0.50 ) 68 kg cos 15 = N. ( )( 9.8 m / s 2 ) ( ) sin( 15 ) (b) The second law equations fo the magnitudes of the x and y components fo the moving cate ae T cosθ f = ma x and T sinθ + N mg = 0. Now f = μkin N. The second equation gives N = mg T sinθ as befoe, so f = μ kin ( mg T sinθ). This expession is substituted fo f in the fist equation to obtain T cosθ μ kin ( mg Tsinθ)= ma x, so the acceleation is a x ( ) = T cosθ + μkin sin θ μ m kin g ( = 304 N )cos sin kg ( ) ( 0.35) ( 9.8 m / s 2 )= 1.3 m/ s The fee-body diagams fo block B and fo the knot just above block A ae shown next. T 1 is the tension foce of the ope pulling on block B o pulling on the knot (as the case may be), T 2 is the tension foce exeted by the second ope (at angle θ = 30 ) on the knot, f is the foce of static fiction exeted by the hoizontal suface on block B, N is nomal foce exeted by the

4 suface on block B, W A is the weight of block A (the magnitude of F gav A ), and W B = 711 N is the weight of block B (the magnitude of F gav B ). m B g m A g Fo each object we take +x hoizontally ightwad and +y upwad. Applying Newton s second law in the x and y diections fo block B and then doing the same fo the knot esults in fou equations: T 1 f stat max = T 1 μ stat N = 0 N W B = 0 T 2 cosθ T 1 = 0 T 2 sinθ W A = 0 whee we assume the static fiction to be at its maximum value (pemitting us to use Eq. 6-11). Solving these equations with μ stat = 0.25, we obtain fo the maximum weight W A = T 2 sinθ = T 1 cosθ sinθ = μstat N cosθ sinθ = μ stat N tanθ = μ stat W B tanθ = 103 N 100 N. 49. The analysis of coodinates and foces (the fee-body diagam) is exactly as in the textbook in TOUCHSTONE EXAMPLE 6-5 (see Fig. 6-30, 6-31, and 6-32). (a) Constant velocity implies zeo acceleation, so the uphill foce must equal (in magnitude) the downhill component of the gavitational foce: T = mg sin θ. Thus, with m = 50 kg and θ = 8.0, the tension in the ope equals 68 N. (b) With an uphill acceleation of 0.10 m/s 2, Newton s second law (applied to the x axis shown in Fig. 6-32(b)) yields fo the magnitude of the tension T mg sin θ = ma x T = mg sin θ + ma x = ( 50 kg)9.8 m / s 2 ( ) sin8.0 + ( 50 kg) 0.10 m / s 2 ( )= 73 N. 51. The solutions to pats (a) and (b) have been combined hee. The fee-body diagam is shown below, with the tension of the sting T, the foce of gavity mg, and the foce of the beeze F. Ou coodinate system is shown. The magnitude of the x-component of the net foce is T sin θ F and the magnitude of the y-component is T cos θ mg, whee θ = 37.

5 mg Since the sphee is motionless the net foce on it is zeo. We answe the questions in the evese ode. Solving T cos θ mg = 0 N fo the tension, we obtain T = mg/cos θ = ( kg) (9.8 m/s 2 ) / cos 37 = N. Solving T sin θ F = 0 N fo the foce of the beeze: F = T sin θ = ( N) sin 37 = N. 57. Fo convenience, we have labeled the 2.0 kg mass m and the 3.0 kg mass M. The +x diection fo m is downhill and the +x diection fo M is ightwad; thus, they acceleate with the same sign. Mg (a) We apply Newton s second law to the magnitude of the components along each block s x axis: mg sin 30 T = ma F + T = Ma Adding the two equations allows us to solve fo the acceleation. mg sin 30 +F = ma + Ma = a(m + M ) a = mg sin 30 +F m + M With F = 2.3 N, we have an acceleation of magnitude a = 1.8 m/s 2. We substitute back in to find the magnitude of the tension T = 3.1 N. (b) We conside the citical case whee the magnitude of F has eached the max value, causing the tension to vanish. The fist of the equations in pat (a) shows that a = g sin 30 in this case;

6 thus, a = 4.9 m/s 2. This implies (along with T = 0 N in the second equation in pat (a)) that magnitude of F = (3.0 kg)(4.9 m/s 2 ) = 14.7 N in the citical case.