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TakeHomeQuizSolution - Picture the Problem The force...

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Unformatted text preview: Picture the Problem: The force vectors acting on blocks A and B as well as the rope knot are shown in the diagram at right. Strategy: Write Newton‘s Second Law for blocks A and B._ as well as Newton’s Second Law for the rope knot. In all cases the acceleration is zero. Combine the equations to solve for E __ which acts on block A and points toward the left. Let the x direction point to the right for block A and down for block B. Solution: 1. (it) ‘Write Newton‘s FI = UK; +11 = E} Second Law for block A: iii-E's ' 2. ‘Write Newton" s Second Law F = -133 + mm? = g for block B: Eluzctlla I 3. “fl-ll}? Newton" s Second Law 2 F; = —TA + 1'; cos 45" = ID for the rope knot: W 25;. = —TE +3; sin45°= 0 mm 4. Divide the Iv equation for the T; Sill 45° _t 455 _ _ 7—3 knot by the x equation: 1: cos 45+: _ an _ _ TA 3. Suhstitute TA = i]; into the a = mgg equation from step 2: ti. Substitute the result into the = TA = mm; = (233 kg}(9_31 {11:51}: 223 N = equation from step 1: .7. {b} As long as mass A is heat-'3' enough that = gramig = [0.320]{8.82 kgMSlEl nu‘s2 = 23".? N 2 22.9 N, the friction force is not affected by changes in mass A. It will if the mass of block A is doubled. In sight: The minimum mass of block A that will satisfy the criteria of step (b) is 129 kg. The answer to (a) is reported with only two significant figures because the angle 45° is only given to two significant figures. 90. Picture the Problem: The free—body diagram of the suitcase is depicted at right. Free-b U Ll} diagram Strategy: Write Newton's Second Law in the vertical and horizontal directions. Because the velocity of the suitcase is constant, the acceleration is zero et-‘erjmrhere. This will produce two equations with two unknowns, the normal force and the strap tension. L'se algebraic substitution to solve the two equations for the unknown variables. Err}. =N+Fsinfl —Ir=o N = mg—Fsind' Solution: 1. {:1} Write Newton’s Second Law in the vertical direction and determine the normal force on the suitcase: I. 1Write Newton's Second Law in the hori- EFI = Fcosfl —;rkN = CI zontal direction and solve for the strap force F: F (“IN _ cos I9 ' - - '/ MEN \' . 3. Substitute the expression for F into the N = mg —| sin :9 equation from step 1: K “305 '9 2' N“ + ,uk taut?) =mg mg {18kg}(9.simr51] JV =—=— 1+,ttr1‘tan6l 1+{0.38]tan45° =128N= 0.13kN {033]{123 N] F: Fri-"t _—= cos 6' _ cos 45° 4. {b} Substitute the normal force into the expression for F fiom step 2: Insight: We bent the rules for significant figures Sligl‘lll)’ in step 4 in order to avoid rounding error. A larger coefficient of fiiction (sag; pk = 0.50 } would require a smaller NH 18 N) but a larger F (83 N). The normal force is smaller because the larger F is pulling upward on the suitcase. supporting more of the weight than before. ...
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TakeHomeQuizSolution - Picture the Problem The force...

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