Dynamics - 5 - Newton's 3rd Law Multi-Body

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High School
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High School Physics
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2024
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 4.6 Newton’s 3rd Law in Multi-Body Problems Recall: Newton’s 3rd Law: ⚫ ⚫ Every action has an equal and opposite reaction Forces exist in pairs Question How does this apply to forces between connected objects (either in direct contact or attached by a string)? Free-Body Diagram ⚫ Consider two blocks (A & B) side-by-side on a flat surface, moving to the right. A B ⚫ If they are pushed from the left with an applied force on A, what are all forces on each block? Free-Body Diagram (as one) Free-Body Diagrams (apart) A B Ex #1 A force is applied to block A, which sits next to block B, pushing both at a constant velocity. If the coefficient of kinetic friction is 0.200, with what force does block A push on block B? Fapp ma 20.0kg mb 10.0kg To solve for Fapp, first must consider connected objects as one. Fnet = 0 Ff FN 30.0kg Fapp = Ff Fg F f = k FN F f = (0.200)(30.0)(9.80) Ff = 58.8N Fapp = 58.8N Fapp Now consider forces on block A alone. Fnet = 0 Fb→a Fapp = F f + Fb→a Ff FN 20.0kg Fapp Fg F f = (0.200)(20.0)(9.80) Ff = 39.2 N Fapp = 58.8N Fb→a = 58.8 − 39.2 = 19.6 N [left] So by Newton’s 3rd Law… Fa →b = 19.6 N [right] Can check this by looking at block B. Fnet = 0 Fa→b = F f FN Fa→b 10.0kg Ff Fg Fa→b = (0.200)(10.0)(9.80) Fa →b = 19.6 N [right] Confirmed result for block A Ex #2 A force of 80.0N is applied to block A, which sits next to block B, accelerating both to the right. If μk = 0.200, with what force does block A push on block B? Fapp ma 20.0kg mb 10.0kg To solve for a, first must consider connected objects as one. Fnet = Fapp − F f Ff FN 30.0kg Fnet = Fapp − k FN Fg Fnet = 80 − (0.200)(30.0)(9.80) Fnet = 21.2 N Fnet 2 a= = 0.7067m / s m Fapp Now consider forces on block A alone. Fnet = ma Fnet = ( 20.0)(0.7067) FN Fb→a Fnet = 14.13N Ff 20.0kg Fapp Fg Fnet = Fapp − F f − Fb→a Fb→a = Fapp − F f − Fnet Fb→a = 80.0 − (0.200)(20.0)(9.80) − 14.13 Fb→a = 26.7 N [left] Fa→b = 26.7 N [right] So by Newton’s 3rd Law… Confirm by checking block B. Fnet = ma Fnet = (10.0)(0.7067) Fnet = 7.067N FN Ff Fa→b 10.0kg Fg Fnet = Fa→b − Ff Fa→b = Fnet + Ff Fa →b = 7.067 + (0.200)(10.0)(9.80) Fa→b = 26.7 N [right] Confirmed result for block A Ex. #3 A 10.0kg mass is suspended by a string & attached to a 5.00kg cart. The force of friction on the system is 40.0N. Mass 5kg Friction = 40N 10kg What is the magnitude of the tension in the string? First, solve for acceleration of system Fnet = Fapplied − Ffriction Fapp = mg Fapp = (10.0)(9.80) Fapp = 98.0N Fnet = 98.0 − 40.0 Fnet = 58.0 N Fnet = ma 58.0 = (15.0)a a = 3.867m / s 2 Now, consider forces on only one object Fnet = Fapplied − Ffriction ma = mg − FT FT 10kg Fg FT = mg − ma FT = (10.0)(9.80) − (10.0)(3.867) FT = 59.3N[up] Can confirm by looking at forces on cart alone FN Fnet = Fapplied − Ffriction ma = FT − F f Ff Mass 5kg FT Fg FT = ma + F f FT = (5.00)(3.867) + 40.0 FT = 59.3N[right] The tension acts equally on hanging mass and cart but in opposite directions (3rd Law).