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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: iiiE's '
2. ‘Write Newton" s Second Law F = 133 + mm? = g
for block B: Eluzctlla I
3. “flll}? 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. Freeb 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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This note was uploaded on 09/02/2011 for the course PHYS 50 taught by Professor Luna during the Spring '11 term at DeAnza College.
 Spring '11
 Luna

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