Correct answer: 0 m/s 2. Explanation: 8 N


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1 Version 001 HW#3  orces rts (00223) 1 his printout should hve 15 questions. Multiplechoice questions my continue on the next column or pge find ll choices before nswering. Angled orce on Block points he horizontl surfce on which the block of mss 5.9 kg slides is frictionless. he force of 37Nctsontheblockinhorizontldirection nd the force of 74 N cts on the block t n ngle s shown below. 37 N 5.9 kg 74 N 60 = 2 cos60 ( ) = ( m ) 1 74 N 37 N 2 = = kg Sttic Equilibrium (prt 1 of 2) 10.0 points he 8 N weight is in equilibrium under the influence of the three forces cting on it. he forcectsfromboveonthelefttnngle of α with the horizontl. he 5.1 N force cts from bove on the right t n ngle of 58 with the horizontl. he force 8 N cts stright down. Wht is the mgnitude of the resulting ccelertion of the block? he ccelertion of grvity is 9.8 m/s 2. α 5.1 N 58 Correct nswer: 0 m/s 2. 8 N Let : 1 = 37 N, 2 = 74 N, α = 60, m = 5.9 kg, nd g = 9.8 m/s 2. Wht is the mgnitude of the force? Correct nswer: N. Stndrd ngulr mesurements re from the positive xxis in counterclockwise direction. 2 α 1 N m mg Let : 1 =, 2 = 5.1 N, α 2 = 58, nd 3 = 8 N. he force 2 hs horizontl component 2 cosα cting to the left, nd the force 1 cts to the right, so net = m = 2 cosα 1 Consider the free body digrm. he green vectors re the components of the slnted forces.
2 Version 001 HW#3  orces rts (00223) 2 α α (prt 1 of 2) 10.0 points Hint: sin 2 θ +cos 2 θ = 1. Consider the 606 N weight held by two cbles shown below. he lefthnd cble hd tension 380 N nd mkes n ngle of θ 2 with the ceiling. he righthnd cble hd tension 430 N nd mkes n ngle of θ 1 with the ceiling. he weight is is equilibrium, so x = 1x + 2x + 3x = 0 1x = 2 cosα 2 0 = (5.1 N)cos58 = N nd y = 1y + 2y + 3y = 0 1y = 2 sinα 2 3 = (5.1 N)sin58 ( 8 N) = N, nd 1 = 1x y = ( N) 2 +( N) 2 = N. 380 N θ N 430 N ) Wht is the ngle θ 1 which the righthnd cble mkes with respect to the ceiling? θ 1 Correct nswer: Observe the freebody digrm below. θ θ (prt 2 of 2) 10.0 points Wht is the ngle α of the force s shown in the figure? Correct nswer: α 1 = rctn = ( 1y 1x ) ( ) N = rctn N mesured from the positive xxis, so α = 180 α 1 = = Hnging Weight 05 W g Note: he sum of the x nd ycomponents of 1, 2, nd W g re equl to zero. Given : W g = 606 N, 1 = 430 N, nd 2 = 380 N. Bsic Concepts: x = 0
3 Version 001 HW#3  orces rts (00223) 3 nd 1 x = x 2 1 cosθ 1 = 2 cosθ 2 (1) 1 2 cos 2 θ 1 = 2 2 cos 2 θ 2 (2) y = 0 y 1 +y 2 +y 3 = 0 1 sinθ sinθ 2 3 = 0 1 sinθ 1 = 2 sinθ sin2 θ 1 = 2 2 sin2 θ sinθ , since (3) 3 sinθ 3 = 3 sin270 = 3, nd 3 cosθ 3 = 3 cos270 = 0. Solution: Since sin 2 θ + cos 2 θ = 1 nd dding Eqs. 2 nd 3, we hve 2 2 = sinθ sinθ 1 = θ 1 = rcsin( ) [ (606 N) 2 +(430 N) 2 = rcsin 2(430 N)(606 N) (380 N) 2 ] 2(430 N)(606 N) = (prt 1 of 2) 10.0 points wo blocks re connected by n extensible mssless cord on n inclined plne s shown in the figure below. he ccelertion of grvity is 9.8 m/s m/s 2 37 kg 63 kg 31 Whtistheforce pullingboththeblocks? Correct nswer: N. Let : θ = 31, M 1 = 37 kg, M 2 = 63 kg, g = 9.8 m/s 2, nd = 3.7 m/s (prt 2 of 2) 10.0 points b) Wht is the ngle θ 2 which the lefthnd cble mkes with respect to the ceiling? N 2 Correct nswer: Using Eq. 1, we hve cosθ 2 = 1 cosθ 1 2 ( ) 1 θ 2 = rccos cosθ 1 ( 2 ) 430 N = rccos 380 N cos = Up Slope N 1 W 1 W 1 W 1 W 2 W 2 θ W 2 he resulting ccelertion is due to the pplied force cting ginst the component of the weights long the inclined surfce. his system is equivlent to combined block of
4 Version 001 HW#3  orces rts (00223) 4 mss M 1 +M 2 being ccelerted up the slope by force, nd cn be described by the eqution (M 1 +M 2 )gsinθ = (M 1 +M 2 ). (1) Solving for we obtin = (M 1 +M 2 )(+gsinθ) = (37 kg+63 kg) ( 3.7 m/s m/s 2) sin31 = N. 007 (prt 2 of 2) 10.0 points Wht is the tension in the cord pulling the lower block? Correct nswer: N. Using Eq. (1) from Prt 1 for the lower block, we hve M 1 gsinθ = M 1. Solving for we obtin = M 1 (+gsinθ) = (37 kg) ( 3.7 m/s m/s 2) sin31 = N. Accelerting Down Plne (prt 1 of 2) 10.0 points A block is relesed from rest on n inclined plne nd moves 2.3 m during the next 2.2 s. he ccelertion of grvity is 9.8 m/s kg µ k 30 Wht is the mgnitude of the ccelertion of the block? Correct nswer: m/s 2. Given : m = 13 kg, l = 2.3 m, θ = 30, nd t = 2.2 s. Consider the free body digrm for the block mg sinθ mg µn N = mgcosθ he ccelertion cn be obtined through kinemtics. Since v 0 = 0, l = v 0 t+ 1 2 t2 = 1 2 t2 = 2l t 2 (1) 2(2.3 m) = (2.2 s) 2 = m/s (prt 2 of 2) 10.0 points Wht is the coefficient of kinetic friction µ k for the incline? Correct nswer: Applying Newton s Second Lw of Motion i = m nd Eq. 1, the sine component of theweight cts down the plne nd friction cts up the plne. he block slides down the plne, so m = mgsinθ µ k mgcosθ 2l ( ) t 2 = g sinθ µ k cosθ 2l = gt 2( ) sinθ µ k cosθ
5 Version 001 HW#3  orces rts (00223) 5 µ k = gt2 sinθ 2l gt 2 (2) cosθ 2l = tnθ gt 2 cosθ = tn30 2(2.3 m) (9.8 m/s 2 )(2.2 s) 2 cos30 = lling Mss (prt 1 of 2) 10.0 points woblocksrerrngedttheendsofmssless cord over frictionless mssless pulley s shown in the figure. Assume the system strts fromrest. Whenthemsseshvemoveddistnce of m, their speed is 1.33 m/s. he ccelertion of grvity is 9.8 m/s 2. 4 kg = (1.33 m/s)2 (0 m/s) 2 2(0.341 m) = m/s 2. Consider free body digrms for the two msses m 2 m 1 µm 2 g m 2 g m 1 g N which leds to 1y : m 1 = m 1 g (1) 2x : m 2 = f k (2) 2y : N = m 2 g, (3) µ 2.6 kg where is the tension in the cord nd from Eq. 3, f k µn = µm 2 g. Becusem 1 ndm 2 retiedtogetherwith cord, they hve sme the speed nd the sme ccelertion. Adding Eqs. 1 & 2 we hve Wht is the coefficient of friction between m 2 nd the tble? Correct nswer: Given : m 1 = 2.6 kg, m 2 = 4 kg, s = m, nd v 0 = 0 m/s. Bsic Concept: Newton s Second Lw = M Solution: heccelertionofm 1 isobtined from the eqution v 2 v 2 0 = 2(s s 0) = v2 v 2 0 2h (m 1 +m 2 ) = m 1 g f k = m 1 g µm 2 g so tht hus µm 2 g = m 1 g (m 1 +m 2 ). µ = m 1 (m 1 +m 2 ) m 2 m 2 g 2.6 kg 2.6 kg+4 kg = 4 kg 4 kg m/s2 9.8 m/s 2 = (prt 2 of 2) 10.0 points Wht is the mgnitude of the tension in the cord? Correct nswer: N.
6 Version 001 HW#3  orces rts (00223) 6 Using Eq. 1 the tension is = m 1 (g ) = (2.6 kg)(9.8 m/s m/s 2 ) = N or, using Eq. 2 nd µ from Prt 1, the tension is = m 2 [+µg] = (4 kg)[( m/s 2 ) +( )(9.8 m/s 2 )] = N. Sincethe isthesmeusing Eqs.1&2: Prt 1, µ = , is verified. Rising wo Msses (prt 1 of 2) 10.0 points wo msses m 1 nd m 2 re connected in the mnner shown. m 1 1 m 2 2 = g 2 he system is ccelerting downwrd with ccelertion of mgnitude g 2. Determine = 1 2 (m 2 m 1 )g 2. 2 = 3 2 m 2g 3. 2 = 1 2 m 2g correct ( ) = 2 m 2 m 1 g 5. 2 = m 2 g ( 6. 2 = m 2 1 ) 2 m 1 g m 2 g m 2 2 = g 2 rom Newton s Second Lw, m 2 g 2 = m 2 g 2. So 2 = m 2 g m 2 g 2 = 1 2 m 2g. 013 (prt 2 of 2) 10.0 points Determine = 5 2 (m 1 +m 2 )g 2. 1 = (m 1 +m 2 )g 3. 1 = 1 2 (m 1 +m 2 )g correct 4. 1 = 2(m 1 +m 2 )g 5. None of these 6. 1 = 3 2 (m 1 +m 2 )g (m 1 +m 2 )g m 1 +m 2 1 = g 2 Consider m 1 ndm 2 swholeobject with mss (m 1 +m 2 ), then So (m 1 +m 2 )g 1 = (m 1 +m 2 ) g 2. 1 = 1 2 (m 1 +m 2 )g. wo Connected Msses 014 (prt 1 of 2) 10.0 points A 14 kg mss onfrictionless inclined surfce is connected to 2.6 kg mss. he pulley is mssless nd frictionless, nd the connecting string is mssless nd does not stretch. he 2.6 kg mss is cted upon by n upwrd force
7 Version 001 HW#3  orces rts (00223) 7 of 3.3 N, nd thus hs downwrd ccelertion of only 5.1 m/s 2. he ccelertion of grvity is 9.8 m/s kg θ block m 2 = m 2 g, so = m 2 g m 2 = (2.6 kg)(9.8 m/s m/s 2 ) 3.3 N = 8.92 N. 5.1 m/s (prt 2 of 2) 10.0 points Wht is the ngle θ? 2.6 kg 3.3 N Wht is the tension in the connecting string? Correct nswer: 8.92 N. Given : m 1 = 14 kg, m 2 = 2.6 kg, = 3.3 N, nd = 5.1 m/s 2. Bsic Concept: i = m i Solution: It is esiest to nlyze the forces on ech block seprtely. Correct nswer: o find the ngle θ, nlyze the freebody digrm for the 14 kg block. You first know tht theccelertionwill hveto be longthe sloped surfce. he tension is prllel to this surfcendsoginsitsfulleffect. heforceof grvity, however, is not prllel to the surfce, to only tht component of the weight tht is prllel to the surfce will contribute to the ccelertion. Note: he weight of m 1 is m 1 g, nd is the sme s before. Hence m 1 = +m 1 gsin(θ), so [ ] θ = sin 1 m1 m 1 g [ (14 kg)(5.1 m/s = sin 1 2 ] ) 8.92 N (14 kg)(9.8 m/s 2 ) = m 2 m 2 g m 1 gsinθ m 1 N m 1 gcosθ Note: isthetensioninthestring, isthe externl force, nd the weight of m 2 is m 2 g. rom the freebody digrm on the 2.6 kg
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