Pulley and Rope Question Determine the unbalanced force and the acceleration of the system, and the tension at each of ( e7 points in the ropes. 2. Kg 4 2.kg 30 MK : O. | FOR ALL SURFACES 2Kg| A E 1. First determine Fg for all hanging masses, FRamp for all masses on a ramp, and Ff for all sliding masses. (Note the masses have been labelled A-E) Vertical Ramp Friction FgA = mg FRB = mgsine Ff8 = umgcose =2(9.8) =10(9.8)sin60 (0.1)(10)(9.8)cos60 =19.6N[Right] =84.87N[Right] =4.9N Fge = mg FRD = mgsine Ffc = umg =1(9.8) =2(9.8)sin30 =(0.1)(2)(9.8) =9.8N[Left] =9.8N[Left] =1.96N . Note: when finding FR or Ff on a ramp FfD = umgcose Always measure @ to the horizontal. =(0.1)(2)(9.8)cos30 =1.70N . Note: Frictional Forces do not have a direction at this point, because we have not determined whether the system is moving to the right or to the left, once that is determined, we know that friction will oppose the motion. 2. Forces Right = FgA + FRB Forces Left = FgE + FRD = 19.6 + 84.87 = 9.8 + 9.8 = 104.47N = 19.6N Since: Forces Right > Forces Left , this tells us that there is an unbalanced force to the Right, and the system will therefore accelerate and move to the right. As a result all Frictional forces must act to the Left.TA = FgE + FRD + FfD + Ffc + ma = 9.8 + 9.8 + 1.70 + 1.96 + 5(4.49) 10' MA "Q. 1 FOR ALL SURFACES = 45.7N[-] TB = FgE + FRD + FfD + ma = 9.8 + 9.8 + 1,70 + 3(4.49) = 34.77N[->] E. (1 15] To = FOA + FRB - FRD - FfB - Ffc - FfD - ma = 19.6+84.87-9.8-4.9-1.96-1.70-16(4.49) = 14.27 JA = Q.1 Fat act T7 = FgE + ma Ta =.9.8 + 1(4.49) 1 19 = 14.29N[~] FOE STEPS INVOLVED! 1. Find Fg for all hanging masses. 2. Find FR for all masses on the ramps. 3. Find Ff for all the sliding masses, 4. Find the direction of motion by totaling Fright and Fleft and determining which is larger. (Do not include any frictional forces at this time) 5. Find Fun , choosing the direction of motion as positive, Include all frictions in this calculation, (they will all be subtracted) 6. Find Acceleration of the system by dividing the Unbalanced Force by the total mass. 7. Find tensions - Note: When solving for the tension in a rope, your answer can be verified if you solve for the tension in the opposite direction. In the sample question above, Ti = Ta, To = T4, and To = Ty Answers may be off by a couple of hundredths of a Newton due to rounding, so best to carry at least 2 decimal places in all answers.THE BIG Tension Problem Determine the acceleration of the system and the tension at each point. Verify 2 of the tensions. HK : 0.15 ON ALL -> - MK = 0 . 1 ON ALL SURFACES - MK = 0.2 ON ALL SURFACES SURFACES 8 kg 6kg 3 kg 4 KG 3 kg 30' 8 kg 5 kg 15 Kg 10 kg 140 4 kq 4 kg 3 kg G 8 kq 60" 2kg 2 ka 50 7kgFun = Forces Right - Forces Left = FgA + FRB - FRD - FgE - FfB - Ffc- FfD = 19.6 + 84.87 - 9.8 - 9.8 - 4.9 - 1,96 - 1,70 = 76.31 N [Right] + This Unbalanced force causes the acceleration of the whole system. mtotal = MA + MB + mc + mD + me = 2+ 10 +2+2+1 = 17kg a = Fun/mtotal = 76.31 / 17 = 4.49 m/s2[Right] Finding Tensions In order to find the tension at any point, imagine the rope breaking at that point, and determine what force is necessary to pull on the broken end of the rope so the blocks continue to accelerate at the same rate. If two points are shown on the same rope, the answers should be the same as each other, with opposite directions. Fun = FgA - T1 2145 T1 = FgA - Fun but, Fun = ma T1 = FgA - MAC F-q T1 = 19.6 - 2(4.49) = 10.62N[up] Fun = T2 + FRB - FfB - Ffc- FRO - FfD - FgE NOT2 = FfB + Fic + FRD + FfD + FgE - FRB + ma 47% =4,9 + 1.96 +9.8 + 1.7 +9.8 - 84.87 + 15(4.49) = 10.64N[~] E [ ikg . Note: the mass used when solving for T2 is 15 kg since mass " A" is not involved, Technically these two answers should have been equal in magnitude due to both Ti and T2 being in the same rope. Our answers differ by 2/100 of a Newton due to the rounding of the acceleration to 4.49m/s2. Without any rounding, you would get . exactly the same answer for both tensions. "T3 = FgA + FRB - Ff8 - ma = 19.6 + 84.87 - 4.9 - 12(4.49) = 45.69N 12 Ky |