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Dynamics Problems Warm Up The system below is in equilibrium. If the scale is calibrated in N, what does it read? Since the tension is distributed over the entire (mass-less) rope, the scale will read (9.8N/kg)(5kg) = 49N 5kg 5kg Warm Up Determine the Normal force in each of the following: N w F N w cos F N W1 W2 F F 1 2 N 0 N F sin mg F F 2 1 m N w cos N mg cos 2 F sin 1 N F sin 1 F sin 2 mg F N F sin Warm Up Write F=ma for each scenario: F F 2 1 m a F ma F cos 1 F cos 2 F sin 1 F sin 2 mg ma F F ma w w sin w cos F a g F 1 F ma 2 mg sin 2 F cos 1 mg cos 2 F sin 1 ma Example 1 A pumpkin of unknown mass is suspended by a cord attached to the ceiling and pushed away from vertical. When a 24.0 N force is applied to the pumpkin at an angle of 18.00 to horizontal, the pumpkin will remain in equilibrium when the cord makes an angle of 32.00 with the vertical. (A) What is the tension in the cord when the pumpkin is in equilibrium ? (B) What is the mass of the pumpkin ? Diagram Free body Diagram FT Fap 24.0N 58 18 Fg (A) What is the tension in the cord when the pumpkin is in equilibrium ? FT Fap 24.0N Since the pumpkin is in static equilibrium, we need only look at the horizontal components Fx 0 FTx Fapx 0 T cos 58 24.0 N cos 18 0 T 24.0 N cos 18 cos 58 43.073N 43.1N 58 18 Fg (B) What is the mass of the pumpkin ? FT Fap 24.0N Since the pumpkin is in static equilibrium, we need only look at the vertical components F y 18 58 0 Fg FTy Fap y Fg 0 T sin 58 Fapp sin 18 mg 0 m T sin 58 Fapp sin 18 g 43.073N sin 58 24.0 N sin 18 9.81 4.4795kg 4.48kg m s2 Example 2 The tension in the horizontal rope is 30N A) Determine the weight of the object Diagram Free body Diagram 500 FT 2 400 30N FT 2 sin 50 500 FT 2 cos 50 FT 1 Fw FT 3 30 N A) Determine the weight of the object FT 2 The weight is in static equilibrium, so the appropriate net component forces must be zero. FT 2 sin 50 500 FT 2 cos 50 FT 1 Fw Horizontal Component F x 0 F T 2x F T 3 0 FT 2 cos 50 FT 3 0 FT 2 cos 50 FT 3 FT 3 FT 2 cos 50 30 N cos 50 46.672 N Vertical Component F y 0 F T 2 y F T1 0 FT 2 sin 50 Fw 0 Fw FT 2 sin 50 46.672 N sin 50 36 N The weight of the mass is 36N FT 3 30 N Weight on a Wire A rope extends between two poles. A 80N weight hangs from it as per the diagram. A) Determine the tension in both parts of the rope. Diagram T1 100 150 T2 80N Free body Diagram T1 T1 sin 15 T2 15 T1 cos 15 10 T2 cos 10 FG T2 sin 10 A) Determine the tension in both parts of the rope. T1 T1 sin 15 The weight is in static equilibrium, so the appropriate net component forces must be zero. T2 15 T1 cos 15 10 T2 cos 10 T2 sin 10 FG Horizontal Component F x Vertical Component F 0 F T 1x F T 2 x 0 T2 cos 10 cos 15 182.8 N cos 10 T1 cos 15 186.4 N 186 N 0 F T1y F T 2 y F G 0 T1 cos 15 T2 cos 10 0 T1 y T1 sin 15 T2 sin 10 FG 0 T2 cos 10 cos 15 sin 15 T2 sin 10 80 N 0 80 N 0.437527 182.8 T2 183 N Example 3 The system in the diagram is just on the verge of slipping. If a 7.0N weight is hanging from a ring, what is the coefficient of static friction between the block and the tabletop?. Free body Diagram Block Ring F Tc FN FTc sin 30 30 35N Ff 7.0N F Tb 30 F Tb FTc cos 30 F Gb FG The system in the diagram is just on the verge of slipping. If a 7.0N weight is hanging from a ring, what is the coefficient of static friction between the block and the tabletop?. The weight is just in static equilibrium, so the appropriate net component forces must be zero. Ring Block F Tc FN Ff 30 F Tb F Tb F Gb From Block: FTb F f f FN f mg F u f Tb mg From ring: FTb FTc cos 30 FG FTc sin 30 FTc mr g sin 30 FG Combining: uf FTb mb g f FTc cos 30 mb g mr g cos 30 sin 30 mb g mr g cos 30 mb g sin 30 7 N cos 30 35 N sin 30 0.35 Example 4 A box with mass 18.0 kg is pulled along the floor. If the horizontal force is 95.0 N [E], what is the acceleration of the box? Pulling a Box (Part 1) A box with mass 18.0 kg is pulled along the floor. If the horizontal force is 95.0 N [E], what is the acceleration of the box? Free body Diagram FN +y FA +x FG a Pulling a Box (Part 1) A box with mass 18.0 kg is pulled along the floor. If the horizontal force is 95.0 N [E], what is the acceleration of the box? FN +y a FA +x FG Vertical Forces F y 0 F N FG 0 N mg 0 N mg Horizontal Forces F x ma x F A ma x FA ma FA m F A m a Solving FA m 95.0 N 18.0kg m 5.28 2 s a Example 5 Two boxes, one with a mass of 4.00 kg and the other with a mass of 6.00 kg are pulled along the floor. If a horizontal force is 20.0 N [E] is applied to the 6.00 kg box, what is the acceleration of each box, and what is the tension T in the rope connecting both boxes? Pulling a Box (Part 2) A two boxes, one with a mass of 4.00 kg and the other with a mass of 6.00 kg is pulled along the floor. If a horizontal force is 20.0 N [E] is applied to the 6.00 kg box, what is the acceleration of each box, and what is the tension T in the rope connecting both boxes? Free body Diagram FN FN FT FA FT a +y +x FG FG Pulling a Box (Part 2) A two boxes, one with a mass of 4.00 kg and the other with a mass of 6.00 kg is pulled along the floor. If a horizontal force is 20.0 N [E] is applied to the 6.00 kg box, what is the acceleration of each box, and what is the tension T in the rope connecting both boxes? FN a FN +y FA FT FT +x Forces 4.00 kg Box F x m1 a x F T m1 a T m1a FG FG 6.00 kg Box F x m2 ax F T F A m2 a T FA m2 a Adding to eliminate T and find a T m1a T FA m2 a + FA m1a m2 a FA a m1 m2 a FA m1 m2 Pulling a Box (Part 2) A two boxes, one with a mass of 4.00 kg and the other with a mass of 6.00 kg is pulled along the floor. If a horizontal force is 20.0 N [E] is applied to the 6.00 kg box, what is the acceleration of each box, and what is the tension T at either end of the rope connecting both boxes? FN a FN +y FT FA FT +x Solve for Acceleration FA a m1 m2 20.0 N 4.00kg 6.00kg m 2.00 2 s FG Now for Tension T m1a N 4.00kg 2.00 kg 8.00 N FG We could have used the other tension formula from Box 2 and obtained the same answer Example 6 A two boxes, one with a mass of 4.00 kg and the other with a mass of 6.00 kg is connected by a 1kg rope and they are pulled along the floor. If a horizontal force is 20.0 N [E] is applied to the 6.00 kg box, what is the acceleration of each box, and what is the tension T at either end of the rope connecting both boxes? 1.00kg Pulling a Box (Part 3) A two boxes, one with a mass of 4.00 kg and the other with a mass of 6.00 kg is pulled along the floor. If a horizontal force is 20.0 N [E] is applied to the 6.00 kg box, what is the acceleration of each box, and what is the tension T at either end of the rope connecting both boxes? Free body Diagram 1.00 kg FN F T1 FG Because the rope has mass, the two ends will experience different tensions FN FA FT 2 a +y +x FG Pulling a Box (Part 3) A two boxes, one with a mass of 4.00 kg and the other with a mass of 6.00 kg is pulled along the floor. If a horizontal force is 20.0 N [E] is applied to the 6.00 kg box, what is the acceleration of each box, and what is the tension T at either end of the rope connecting both boxes? a FN FN +y FT FA FT +x Forces 4.00 kg Box F x m1 a x F T m1 a T1 m1a FG 6.00 kg Box F x m2 ax F T F A m2 a T2 FA m2 a T2 FA m2 a FG Using F=ma for the system to find a F ma F m m A a 1 2 mrope a FA m1 m2 mrope Pulling a Box (Part 3) A two boxes, one with a mass of 4.00 kg and the other with a mass of 6.00 kg is pulled along the floor. If a horizontal force is 20.0 N [E] is applied to the 6.00 kg box, what is the acceleration of each box, and what is the tension T in the rope connecting both boxes? FN FN a +y FT FA FT +x FG FG Solve for Acceleration a FA m1 m2 mrope 20.0 N 4.00kg 1.00kg 6.00kg m 1.82 2 s Now for T1 T1 m1a N 4.00kg 1.82 kg 7.27 N Now for T2 T2 FA m2 a N 20.0 N 6.00kg 1.82 kg 9.09 N Example 7 A worker drags a 38.0 kg box along the floor by pulling on a rope attached to the box. The coefficient of friction between the floor and the box is us=0.450 and uk=0.410. a) What are the force of friction and acceleration when the worker applies a horizontal force of 150N? b) What are the force of friction and acceleration when the worker applies a horizontal force of 190N? 38kg Friction 1: Solution (Free Body Diagram) What are the force of friction and acceleration of the worker applies a horizontal force of 150N? The Normal force up Friction to the left FN The applied force of tension to the right 38kg FT Ff FG The force of gravity down +y +x Example 7a: Solution (Vector Components) What are the force of friction and acceleration of the worker applies a horizontal force of 150N? FN +y 38kg FT Ff +x FG To determine if the box will move, we must find the maximum static friction and compare it to the applied force. F y ma y 0 Ff S FN F N FG 0 0.450 372.4 N FN mg 0 168 N FN mg N 38kg 9.80 kg 372.4 N Since the applied force by the worker is only 150N, the box will not move Example 7b: Solution (Vector Components) What are the force of friction and acceleration of the worker applies a horizontal force of 190N? FN +y 38kg FT Ff +x FG x ma x Since applied force is greater than 168N from part a), we will have an Fthe nd acceleration F f F T ma x in the x direction. So we will apply Newton’s 2 Law in the horizontal direction. F F ma F =(0.410)(372.4N)=153N f T x a K FT Ff m 190 N K FN m 190 N 0.410 372.4 N 38kg m 0.982 2 s The acceleration of the box is 0.982 m/s2 [E] Example A train of three masses is pulled along a frictionless surface. Calculate the tensions in the ropes. T1 8 kg 30 N T2 5 kg 13 kg We can find the acceleration of the train by treating the three masses as one unit. F ma 30 N 8kg 5kg 13kg a a 1.15 F ma m [left ] 2 s Tension in rope T1 T1 Tension in rope T2 F F ma FA T1 m1a T1 FA m1a m 30 N 8kg 1.15 2 s 20.8 N T1 T1 T2 m2 a or T2 T1 m2 a m 20.8 N 5kg 1.15 2 s 15.1N T2 F ma T2 ma m 13kg 1.15 2 s 15.1N Example 8 A farmer uses his tractor to pull a sled loaded with firewood. Suppose the chain pulls the sled at and angle of 300 above the horizontal. How hard does the tractor have to pull to keep the sled moving with a constant velocity if the sled and firewood has a weight of 15,000 N and uk=0.40? 300 Example 8: Solution (Free Body Diagram) A farmer uses his tractor to pull a sled loaded with firewood. Suppose the chain pulls the sled at and angle of 300 above the horizontal. How hard does the tractor have to pull to keep the sled moving with a constant velocity if the sled and firewood has a weight of 15,000 N and uk=0.40? The Normal force up Friction to the left FN The applied force of tension at 300 FT FT sin 30 30 FT cos 30 Ff 300 Tension broken down into components FG The force of gravity down Example 8: Solution (Force Components) A farmer uses his tractor to pull a sled loaded with firewood. Suppose the chain pulls the sled at and angle of 300 above the horizontal. How hard does the tractor have to pull to keep the sled moving with a constant velocity if the sled and firewood has a weight of 15,000 N and uk=0.40? FN FT +y FT sin 30 30 FT cos 30 Ff +x 300 FG Vertical Components Horizontal Components Fy 0 F x ma x F N F G F Ty 0 F f F Tx 0 FN mg FT sin 30 0 k FN FT cos 30 0 FN mg FT sin 30 Since we have a constant velocity, acceleration is 0 uk mg FT sin 30 FT cos 30 0 We will need FN, so solve for FN Example 8: Solution (Force Components) A farmer uses his tractor to pull a sled loaded with firewood. Suppose the chain pulls the sled at and angle of 300 above the horizontal. How hard does the tractor have to pull to keep the sled moving with a constant velocity if the sled and firewood has a weight of 15,000 N and uk=0.40? FN FT +y FT sin 30 30 FT cos 30 Ff +x 300 FG Solve for FT uk mg FT sin 30 FT cos 30 0 uk mg uk FT sin 30 FT cos 30 0 uk FT sin 30 FT cos 30 uk mg FT uk sin 30 cos 30 uk mg FT uk mg uk sin 30 cos 30 FT 0.40 15000 N 0.40 sin 30 cos 30 5628.384 N 5600 N Example 9 A group of children toboggan down a hill with a 300 slope. Given that the coefficient of kinetic friction is 0.10, calculate their acceleration and the speed they will obtain after 6.0s. Example 9: Solution (Free Body Diagram) A group of children toboggan down a hill with a 300 slope. Given that the coefficient of kinetic friction is 0.10, calculate their acceleration and the speed they will obtain after 6.0s. +y Choose axis orientation to match the direction of motion Normal is and the normal to the surface perpendicular to the surface +x FN Ff Object Force of friction F cos 30 opposes direction of motion G FN Ff 30 FG sin 30 Decompose gravity into axis components FG cos 30 30 FG sin 30 FG Force of gravity is straight down FG Example 9: Solution (Force Vectors) A group of children toboggan down a hill with a 300 slope. Given that the coefficient of kinetic friction is 0.10, calculate their acceleration and the speed they will obtain after 6.0s. FN +y F f FG cos 30 30 +x FG sin 30 FG y direction F y 0 F N F Gy 0 FN FG cos 30 0 FN FG cos 30 FN mg cos 30 x direction F max Remember Ff to FGxsolve max for FN because we uk F FG sin 30 max will need itNlater uk mg cos 300 mg sin 30 max x ax uk g cos 300 g sin 30 Example 9: Solution (Force Vectors) A group of children toboggan down a hill with a 300 slope. Given that the coefficient of kinetic friction is 0.10, calculate their acceleration and the speed they will obtain after 6.0s. FN +y F f FG cos 30 30 +x FG sin 30 FG Acceleration ax uk g cos 30 g sin 30 0 m m 0.10 9.80 2 cos 30 9.80 2 sin 30 s s m 4.05 2 s m 4.1 2 s Speed v f vi at m m 4.05 2 6s s s m 24.3 s m 24 s 0 Example 10 Block A in the diagram below weighs 1.50 N and block B weighs 4.30 N. The coefficient of kinetic friction between all surfaces is 0.20. Find the magnitude of the horizontal force, F, necessary to drag block B to the left at a constant speed if A and B are connected by a light, flexible cord passing around a fixed, frictionless pulley. Example 10: Solution (Free Body Diagram) Block A in the diagram below weighs 1.50 N and block B weighs 4.30 N. The coefficient of kinetic friction between all surfaces is 0.20. Find the magnitude of the horizontal force, F, necessary to drag block B to the left at a constant speed if A and B are connected by a light, flexible cord passing around a fixed, frictionless pulley. +y Box B FN a +x FN Normal Force Friction from Friction from Object B A Table F f (box A ) F f (box A ) F f ( table ) Force of Gravity F T A and B from FA Applied Force FG F f ( table ) FA FT Tension FG Example 10: Solution (Free Body Diagram) Block A in the diagram below weighs 1.50 N and block B weighs 4.30 N. The coefficient of kinetic friction between all surfaces is 0.20. Find the magnitude of the horizontal force, F, necessary to drag block B to the left at a constant speed if A and B are connected by a light, flexible cord passing around a fixed, frictionless pulley. FN +y a +x Normal Force Box A F f ( box A ) F f ( table ) FA FT FN FG FN Friction Object Tension FT Ff FT Ff F GA +y a +x Force of Gravity from A only F GA Example 10: Solution (Force Vectors) Block A in the diagram below weighs 1.50 N and block B weighs 4.30 N. The coefficient of kinetic friction between all surfaces is 0.20. Find the magnitude of the horizontal force, F, necessary to drag block B to the left at a constant speed if A and B are connected by a light, flexible cord passing around a fixed, frictionless pulley. A B FN FN a +y F f ( box A ) FT Ff +x F f ( table ) FA FT F GA FG x-direction for Box A F x 0 F f FT 0 F f FT 0 k FN FT 0 k mA g FT 0 FT k mA g x-direction for Box B F x 0 FA F f (box A) F f (table ) F T 0 FA k FN ( A) k FN (table ) FT 0 FA k mA g k mA mB g k mA g 3k mA g uk mB g FA k mA g k mA mB g FT 0 FA k mA g k mA mB g FT This was determined using Box A Example 10: Solution (Force Vectors) Block A in the diagram below weighs 1.50 N and block B weighs 4.30 N. The coefficient of kinetic friction between all surfaces is 0.20. Find the magnitude of the horizontal force, F, necessary to drag block B to the left at a constant speed if A and B are connected by a light, flexible cord passing around a fixed, frictionless pulley. A B FN FN a +y F f ( box A ) FT Ff +x F f ( table ) FA FT F GA FG FA 3k mA g uk mB g 3 0.20 1.50 N 0.20 4.30 N 1.76 N Example 11 How much force is needed to give the blocks an acceleration of 3.0 m/s 2 if the coefficient of kinetic friction between the blocks and the floor is 0.20 ? How much force does the 1.50 kg block exert on the 2.0 kg block? 2.0 1.5 kg kg 1.0 kg Example 11a (free body diagram) How much force is needed to give the blocks an acceleration of 3.0 m/s 2 if the coefficient of kinetic friction between the blocks and the floor is 0.20 ? How much force does the 1.50 kg block exert on the 2.0 kg block? 3 Blocks taken as a Single Unit Normal Force FN Friction Object Applied FA Ff FG +y a +x Force of Gravity 2.0 1.5 kg kg 1.0 kg Example 11a (force vectors) How much force is needed to give the blocks an acceleration of 3.0 m/s 2 if the coefficient of kinetic friction between the blocks and the floor is 0.20 ? How much force does the 1.50 kg block exert on the 2.0 kg block? FN +y 1.5 kg a FA Ff 2.0 kg 1.0 kg +x FG F x max Ff FA ma Ff FA ma k FN FA ma mg FA ma FA ma mg m N 1.5kg 2.0kg 1.0kg 3.0 2 0.20 1.5kg 2.0kg 1.0kg 9.8 s kg 22.32 N 22 N Example 11b (free body diagram) How much force is needed to give the blocks an acceleration of 3.0 m/s 2 if the coefficient of kinetic friction between the blocks and the floor is 0.20 ? How much force does the 1.50 kg block exert on the 2.0 kg block? 2.0 kg Block and 1.0 kg taken as a Single Unit Normal Force Since we are considering 2.0kg and 1.0kg block as a unit, then the Force is the push of 1.5 kg block on the combined block FN Friction Object Applied FA Ff Force of Gravity F G2.01.0 +y a +x 2.0 1.5 kg kg 1.0 kg Example 11b (force vectors) How much force is needed to give the blocks an acceleration of 3.0 m/s 2 if the coefficient of kinetic friction between the blocks and the floor is 0.20 ? How much force does the 1.50 kg block exert on the 2.0 kg block? +y FN +x F block1 Ff F G2.01.0 F x a 2.0 kg 1.0 kg ma x F f F block 1 ma uFN Fblock 1 ma umg Fblock 1 ma Fblock 1 ma mg m N 2.0kg 1.0kg 3.0 2 0.20 2.0kg 1.0kg 9.8 s kg 14.88 N 15 N Example 12 Blocks A, B, and C are placed as in the figure and connected by ropes of negligible mass. Both A and B weigh 25 N each, and the coefficient of kinetic friction between the blocks and the surface is 0.35. Block C descends with constant velocity. a) Determine the tension in the rope connecting Block A and B. b) What is the weight of Block C? c) If the rope connecting A and B were cut, what would be the acceleration of C? Example 12 Blocks A, B, and C are placed as in the figure and connected by ropes of negligible e mass. Both A and B weigh 25 N each, and the coefficient of kinetic friction between the blocks and the surface is 0.35. Block c descends with constant velocity. a) Determine the tension in the rope connecting Block A and B. b) What is the weight of Block C? c) If the rope connecting A and B were cut, what would be the acceleration of C? FN F T (C ) FN Ff F F Tf F T ( A) Normal FG FN Object Ff Friction FG FG Block B Normal Block A Tension from C FN Tension FT Friction Tension from A Ff Gravity F T ( A) F T (C ) Object 36.9 FG Gravity Example 12 (Force Vectors) Blocks A, B, and C are placed as in the figure and connected by ropes of negligible e mass. Both A and B weigh 25 N each, and the coefficient of kinetic friction between the blocks and the surface is 0.35. Block c descends with constant velocity. a) Determine the tension in the rope connecting Block A and B. b) What is the weight of Block C? c) If the rope connecting A and B were cut, what would be the acceleration of C? FN FN Ff F T (C ) mB g cos 36.9 FT Ff 36.9 mB g sin 36.9 F T ( A) FG FG Block A F x 0 F f ( A) F T ( A) 0 k FN FT A 0 k mA g FT A 0 Block B F x 0 F f B F T A F Gx F T C 0 k FN B FT A mB g sin 36.9 FT C 0 k mB cos 36.9 g FT A mB g sin 36.9 FT C 0 Block C FT C mC g Example 12 (Force Vectors) Blocks A, B, and C are placed as in the figure and connected by ropes of negligible e mass. Both A and B weigh 25 N each, and the coefficient of kinetic friction between the blocks and the surface is 0.35. Block c descends with constant velocity. a) Determine the tension in the rope connecting Block A and B. b) What is the weight of Block C? c) If the rope connecting A and B were cut, what would be the acceleration of C? FN Ff FN F T (C ) mB g cos 36.9 FT Ff 36.9 mB g sin 36.9 F T ( A) FG Block A k mA g FT A 0 FT A k mA g 0.35 25 N 8.75 N 8.8 N FG Block C Block B k mB cos 36.9 g FT A mB g sin 36.9 FT C 0 FT C mC g k mB cos 36.9 g FT A mB g sin 36.9 mC g 0 mC g k mB cos 36.9 g FT A mB g sin 36.9 Solving for25the 0.35 N cos 36.9 8.75 N 25 N sin 36.9 Applying Block tension between C’s equation to 30.757 N The weight block A and B Block of Block C B 31N Example 12 (Force Vectors) Blocks A, B, and C are placed as in the figure and connected by ropes of negligible e mass. Both A and B weigh 25 N each, and the coefficient of kinetic friction between the blocks and the surface is 0.35. Block c descends with constant velocity. a) Determine the tension in the rope connecting Block A and B. b) What is the weight of Block C? c) If the rope connecting A and B were cut, what would be the acceleration of C? FN FN Ff F T (C ) mB g cos 36.9 FT Ff 36.9 mB g sin 36.9 F T ( A) FG FG When the rope is cut: F ma mC g Ff B mB g sin 36.9 mB mC a a ) FT A 8.8 N b) mC g 31N c) a 1.5 m s2 a mC g Ff B mB g sin 36.9 mB mC 30.7 N 0.35 25 N cos 36.9 25 N sin 36.9 2.55kg 3.14kg m 1.53 2 s Example 13 A small rubber stopper is hung from the rear view mirror in a truck. The cord makes an angle of 9.20 with the vertical as the truck is accelerating forward. Determine the magnitude of the acceleration. 9.2 Example 13 (Free Body Diagram) A small rubber stopper is hung from the rear view mirror in a truck. The cord makes an angle of 9.20 with the vertical as the truck is accelerating forward. Determine the magnitude of the acceleration. We will look at this from outside the truck (ie the ground) because we would prefer an inertial frame of reference. Tension broken into components Tension FT FT FT sin 9.2 FT cos 9.2 9.2 Object 9.2 Gravity FG FG Example 13 (Force Vectors) A small rubber stopper is hung from the rear view mirror in a truck. The cord makes an angle of 9.20 with the vertical as the truck is accelerating forward. Determine the magnitude of the acceleration. FT sin 9.2F T FT cos 9.2 9.2 FG Vertical Forces F F x y 0 F G F Ty 0 mg FT cos 9.2 0 FT Horizontal Forces mg cos 9.2 max FT sin 9.2 max FT sin 9.2 m sin 9.2 mg cos 9.2 m ax g sin 9.2 cos 9.2 ax g tan 9.2 m 9.8 2 tan 9.2 s m 1.6 2 s Example 14 Calculate the acceleration of a box that experiences a magnetic force of repulsion of 12N from a wall and a coefficient of friction of 0.40 between the wall and the box. The box is also being pushed against the wall with a force of 25N at 300 to the horizontal. 5.0 kg 25N 30 Example 14 (Free Body Diagram) Calculate the acceleration of an box that experiences a magnetic force of repulsion of 12N from a wall and a coefficient of friction of 0.40 between the wall and the box. The box is also being pushed against the wall with a force of 25N at 300 to the horizontal. Applied Friction Object 25sin 30 Magnetic Since the gravity force down (5x9.8) is greater than force up (25sin(30), the box slides down, so friction is up. 5.0 kg 25cos 30 Normal 25N 30 Gravity Example 14 (Vector Forces) Calculate the acceleration of an box that experiences a magnetic force of repulsion of 12N from a wall and a coefficient of friction of 0.40 between the wall and the box. The box is also being pushed against the wall with a force of 25N at 300 to the horizontal. 25sin 30 +x 5.0 kg a 25cos 30 +y Horizontal F x 0 F Ax F M F N 0 FAx FM FN 0 FN FAx FM FN 25 N cos 30 FM Vertical F y 25N 30 ma y F Ay F f F G ma y FAy Ff FG ma y ay FAy Ff FG m 25 N sin 30 k FN mg m 25 N sin 30 k 25 N cos 30 FM mg m Example 14 (Insert Numbers) Calculate the acceleration of an box that experiences a magnetic force of repulsion of 12N from a wall and a coefficient of friction of 0.40 between the wall and the box. The box is also being pushed against the wall with a force of 25N at 300 to the horizontal. 25sin 30 +x 5.0 kg a 25cos 30 +y 25N 30 ay 25 N sin 30 k 25 N cos 30 FM mg m m 25 N sin 30 0.4 25 N cos 30 12 N 5kg 9.8 2 s 5.0kg m 6.5 2 s Example 15 Three hundred identical objects are connected in series on a ramp. Calculate the force between objects 120 and 121 if each object has a mass of 0.50 kg and the applied force that is pulling them has a magnitude of 300N. The objects are accelerating at 0.88 m/s 2, and the coefficient of kinetic friction is 0.15. Example 15 (Free Body Diagram) Three hundred identical objects are connected in series on a ramp. Calculate the force between objects 120 and 121 if each object has a mass of 0.50 kg and the applied force that is pulling them has a magnitude of 300N. The objects are accelerating at 0.88 m/s 2, and the coefficient of kinetic friction is 0.15. For Masses 121 to 300 Treat objects 300 thru 121 as a single mass of 180*0.5kg Normal Big Mass FN F T ( above ) Ff 10 Tension mg cos 10from above Gravity Friction FG mg sin 10 Example 15 (Vector Forces) Three hundred identical objects are connected in series on a ramp. Calculate the force between objects 120 and 121 if each object has a mass of 0.50 kg and the applied force that is pulling them has a magnitude of 300N. The objects are accelerating at 0.88 m/s 2, and the coefficient of kinetic friction is 0.15. For Combined Masses 121 to 300 FN F T ( above ) a +y +x Ff 10 FG y-axis mg cos 10 mg sin 10 x-axis F y 0 F N F Gy 0 FN FGy 0 FN mg cos 10 0 FN mg cos 10 F x ma x F f F Gx F T ma x Ff FGx FT ma FN mg sin 10 FT ma FT ma FN mg sin 10 FT ma mg cos 10 mg sin 10 Example 15 (Insert Values) Three hundred identical objects are connected in series on a ramp. Calculate the force between objects 120 and 121 if each object has a mass of 0.50 kg and the applied force that is pulling them has a magnitude of 300N. The objects are accelerating at 0.88 m/s 2, and the coefficient of kinetic friction is 0.15. For Combined Masses 1 to 120 FN F T ( above ) a +y +x Ff 10 FG mg cos 10 mg sin 10 FT ma mg cos 10 mg sin 10 m m m 180 0.5kg 0.88 2 0.15 180 0.5kg 9.8 2 cos 10 180 0.5kg 9.8 2 sin 10 s s s 362.6477 N 360 N Example 16 A person exerts a force of 175N at W300S on a 20.0 kg crate which slides to the left across a level floor when u=0.400, a) Find the normal force on the crate b) Find the force of friction on the crate c) Find the acceleration of the crate. 300 a Example 16 (Free Body Diagram) A person exerts a force of 175N at W300S on a 20.0 kg crate which slides to the left across a level floor when u=0.400, a) Find the normal force on the crate b) Find the force of friction on the crate c) Find the acceleration of the crate. FN a Ff 300 +y Fa Fg +x 300 Example 16 (Vector Forces) A person exerts a force of 175N at W300S on a 20.0 kg crate which slides to the left across a level floor when u=0.400, +y FN a) Find the normal force on the crate b) Find the force of friction on the crate +x c) Find the acceleration of the crate. a Ff 300 y-axis F Fa y 0 F N F Gy F Ay 0 FN FGy FAy 0 FN mg FA sin 30 0 FN mg FA sin 30 Insert Values Fg m FN 20.0kg 9.8 2 175 N sin 30 s 196 N 87.5 N 283.5 N 284 N Example 16 (Vector Forces) A person exerts a force of 175N at W300S on a 20.0 kg crate which slides to the left across a level floor when u=0.400, FN +y a) Find the normal force on the crate b) Find the force of friction on the crate c) Find the acceleration of the crate. +x a Ff 300 Friction Fa F f k FN F f 0.400 283.5 N F f 113.4 N Ff 113N Fg Example 16 (Vector Forces) A person exerts a force of 175N at W300S on a 20.0 kg crate which slides to the left across a level floor when u=0.400, FN +y a) Find the normal force on the crate b) Find the force of friction on the crate c) Find the acceleration of the crate. +x a Ff 300 Acceleration F x max F f F Ax max Ff FA cos 30 max 113.4 N 175N cos 30 max max 38.1544 N Fa 38.1554 N 20.0kg m ax 1.9077 2 s m ax 1.91 2 W s ax Fg Example 17 Calculate the unknowns for each accelerated block. a) 3.7 m s2 18 kg b) F 0.2 F ma FA Ff ma FA ma Ff FA ma FN m N FA 18kg 3.7 2 0.2 18kg 9.8 s kg FA 102 N c) a 6 kg 0.2 40N 41N 5.6 m s2 m 0.3 Example 17 (Solution) Calculate the unknowns for each accelerated block. a) 3.7 m s2 18 kg F 0.2 F ma FA Ff ma FA ma Ff FA ma FN m N FA 18kg 3.7 2 0.2 18kg 9.8 s kg FA 102 N Example 17 (Solution) Calculate the unknowns for each accelerated block. b) a 6 kg 40N 0.2 F ma FA Ff ma a FA Ff m F FN a A m N 40 N 0.2 6kg 9.8 kg a 6kg a 4.7 m s2 Example 17 (Solution) Calculate the unknowns for each accelerated block. c) 5.6 41N m s2 m 0.3 F ma FA mg ma m a g FA m m FA a g 41N m m 5.6 0.3 9.8 s2 s2 m 4.8kg Example 18 Determine the acceleration of the cart that is required to prevent Sir Isaac Newtant from falling. The coefficient of static friction between the block and Newtant is μs Example 18 Determine the acceleration of the cart that is required to prevent Sir Isaac Newtant from falling. The coefficient of static friction between the block and Newtant is μs If I do not fall, then the friction force, Ff, must balance my weight mg, that is Ff = mg The only horizontal force is the Normal Force .Therefore F=N=ma Putting this together we obtain: Ff s N s ma mg s ma g s a a g s