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PHYSICS SEMESTER ONE UNIT 4: ANSWERS TO PROBLEMS UNIT 4: ANSWERS TO PROBLEMS PROBLEM 1 Consider the following diagram showing the two horizontal forces applied by two friends as they pull a 1.20 × 103 kg car along an icy road. a) Find the total force (in component form). +y F1 = 456 N This is just another vector problem. We have to add two vectors just like we would with displacement vectors. The only difference is the units are N instead of m. F2 = 642 N 2 = 50.0° 75.0° +x The variables are defined in the diagram that came with the question. We need to find θ1 measured from the positive x-axis in order to find the components of F1 . +y Using the geometry θ F1 θ1 = 180.0 º – 50.0º = 130.0° F1y 50.0° The components are then +x F1x F1 y F1 sin 1 456 N sin130.0 349 N F1x F1 cos 1 = 456 N cos130.0 293 N +y F2 y F2 sin 2 642 N sin 75.0 620 N F2 x F2 cos 2 642 N cos 75.0 166 N F2 F1y 75.0° F1x +x The sum of the forces F12 is Fnet F1 F2 293 N ˆi 349 N ˆj 166 N ˆi 620 N ˆj 293 N 166 N ˆi 349 N 620 N ˆj 127 N ˆi 969 N ˆj The net force is Fnet 127 N ˆi 969 N ˆj . If the question had not specified the form of the force vector, we would have continued to calculate the angle and magnitude of the vector to get an answer in the same form as the original question. Creative Commons Attribution 3.0 Unported License 1 PHYSICS SEMESTER ONE UNIT 4: ANSWERS TO PROBLEMS b) Calculate the acceleration (magnitude and direction) of the car. Besides the forces from part a), we have m 1.20 103 kg, a ? Newton’s second law of motion states Fnet ma Rearranging a Fnet m 127 N ˆi 969 N ˆj 1.20 103 kg 127 N ˆi 1.20 103 kg 969 N ˆj 1.20 103 kg 0.1058 m/s 2 ˆi 0.8075 m/s 2 ˆj Converting to magnitude – direction form a 0.1058 m/s 0.8075 m/s 2 2 2 2 0.814 m/s 2 0.8075 m/s 180 arctan 0.1058 m/s 2 2 82.5 180 add 180° because x component is negative 97.5 The car accelerates at 0.81 m/s2 at 97.5° from the x-axis Creative Commons Attribution 3.0 Unported License 2 PHYSICS SEMESTER ONE UNIT 4: ANSWERS TO PROBLEMS PROBLEM 2 Earth is attracted towards that 70 kg man with a force of 686 N. Why don’t we notice the effect of our gravitational force on Earth? Define Fg = 686 N, and the mass of the earth M = 5.981024 kg. If the noted force was the only force acting on the earth, the acceleration of the earth would be a Fg M 686 N 5.98 1024 kg 1.1 1022 m/s 2 This is tiny. Starting at rest and accelerating at this rate for the entire age of the universe (~15 billion years) would result in a final velocity of 510-5 m/s. When we include all of the other forces on the earth (moon, sun, planets, galaxy, other people …), the force from one person is just a tiny part of the force on the earth. The gravitational force from the Earth on us is large compared to most of the major forces acting on us. The gravitational force from us on the Earth on us is small compared to most of the major forces acting on the earth. Creative Commons Attribution 3.0 Unported License 3 PHYSICS SEMESTER ONE UNIT 4: ANSWERS TO PROBLEMS PROBLEM 3 Problem 3 and Page 12 Normal and Gravitational Forces on Figures Draw the normal and gravitational forces in the following systems. FN FN surface FN Fg Fg surface surface Fg Problem 3 Find the acceleration of a 1600 kg car that is being pushed horizontally with a force of 5200 N to the right, assuming that the coefficient of friction is 0.25. Is there acceleration? We expect acceleration to the right so we set the x-axis along that direction, with the y-direction vertical. y FN x Fk Draw FBD, define directions v Define terms: m 1600 kg, F 5200 N ˆi, k 0.25, Fg F a ? We have forces in all directions. The normal force is just enough to balance the force of gravity so there is no acceleration in the y-direction. The equation for the forces in the x-direction is Fnet , x F Fk max ax F Fk m Creative Commons Attribution 3.0 Unported License 4 PHYSICS SEMESTER ONE UNIT 4: ANSWERS TO PROBLEMS Here, the friction force will be against the direction of motion so the sign for the friction force is included in the formula (the alternative is to add the friction force and give it a negative value, both methods are valid). In this expression, we know mass and applied force so we need to find a way to get the friction force. We know that Fk k FN and we can use the motion in the y-direction to find FN . Fnet , x FN Fg 0 FN Fg mg 1600 kg 9.8 m/s 2 1.568 104 N Now, we can find the friction force. Fk k FN 0.25 1.568 10 4 N 3.92 103 N The acceleration is then ax F Fk m 5200 N 3920 N 1600 kg 0.80 m/s 2 The acceleration is 0.80 m/s2 to the right (positive x-direction). Creative Commons Attribution 3.0 Unported License 5 PHYSICS SEMESTER ONE UNIT 4: ANSWERS TO PROBLEMS PROBLEM 4 A boat is secured to a lakeside pier with two horizontal ropes. A wind is blowing off shore. The tension in the ropes are T1 = 48 N [16° N of E] and T2 = 48N [16° S of E]. Assuming that the net horizontal force on the boat is zero, what is the force of the wind on the boat? There is no movement or tension in the vertical direction so we will only look at the horizontal (E-N) plane. T1 Draw FBD, draw directions Fw 16° -16° Terms defined in question. Is the object accelerating? N no What is the net force on the boat? E T2 0 The sums of the force components in both directions must be . 0 Break the tension forces into components T1E T1 cos 1 T2 E T2 cos 2 48 N cos16 48 N cos 16 46.1 N 46.1 N T1N T1 sin 1 T2 N T2 sin 2 48 N sin16 48 N sin 16 13.2 N 13.2N Use the net force to find the components of the wind force Fw . Fnet T1 T2 Fw 0 Rearranging Fw T1 T2 FwE T1E T2 E FwN T1N T2 N 46.14 N 46.14 N 13.2 N 13.2 N 93.3 N 0N The force from the wind is –93 N to the east, or 93 N to the west. Creative Commons Attribution 3.0 Unported License 6 PHYSICS SEMESTER ONE UNIT 4: ANSWERS TO PROBLEMS PROBLEM 5 Problem 5 Imagine, instead of pushing the car in Problem 3, you are tow the same car (mass 1600 kg, coefficient of friction 0.25) with a tow line that makes an angle of 37° with the horizontal. The tension on the tow line is 5200 N. What is the acceleration of the car? y Draw the FBD Define terms: FN x T m 1600 kg, T 5200 N ˆi 37 , k 0.25, Fk ax ? θ = 37° v Once again, we expect the acceleration to be in the x-direction only (any acceleration in the y-direction would drive the car off the ground or into the ground). The big difference from problem 3 is the tension with components in both x and y-directions. All other forces are in single directions. Fg The net force in the x-direction is Fnet , x Tx Fk max Rearranging, ax T cos Fk , m We need to use the y-direction net force equation to find the normal force, and use that to find the force of friction. Fnet , x FN Ty Fg 0 FN Fg Ty mg T sin 1600 kg 9.8 m/s 2 5200 kg cos 37 1.255 104 N Now, we can find the friction force. Creative Commons Attribution 3.0 Unported License 7 PHYSICS SEMESTER ONE UNIT 4: ANSWERS TO PROBLEMS Fk k FN 0.25 1.255 10 4 N 3.14 103 N The acceleration is then ax T cos Fk m 5200 N cos 37 3140 N 1600 kg 0.63 m/s 2 The acceleration is 0.63 m/s2 to the right (positive x-direction). Creative Commons Attribution 3.0 Unported License 8 PHYSICS SEMESTER ONE UNIT 4: ANSWERS TO PROBLEMS PROBLEM 6 Problem 6 – Hint A child pulls a sled up a snow-covered hill at a constant velocity with a force parallel to the hillside. If the sled has a mass of 9.5 kg, the hill slope is 12°, and the coefficient of friction between the sled and the snow is 0.20, what is the force applied by the child? There are several important notes to solving this problem. Check out the hints. y FN x F Draw FBD Fk Hint 1: The normal force is always perpendicular to the surface. In this case, it is not opposite the force of gravity. v θ = 12° Fgy Fg Fgx Hint 2: Choose directions as shown with the x in the direction of motion (up the ramp) and y in the direction of the normal. This greatly simplifies the math because we only have to break the force of gravity into components. The other forces all lie entirely along the x or y in the direction. We will need to find the components of the gravitational force (see the diagram to the right to verify that the angle between the y-direction and gravitational force vector is 12°). y x θ = 12° Fgy mg cos Fgx mg sin Hint 3: Notice that the trig functions usually associated with the x and y- components are switched. This comes from the fact that the angle is measured relative to the y-axis instead of the x-axis. θ = 12° Fgy Fg Fgx The set of hints here are steps common by all ramp problems. Ramp problems show up on assignments and exams so become very familiar with the steps. Creative Commons Attribution 3.0 Unported License 9 PHYSICS SEMESTER ONE UNIT 4: ANSWERS TO PROBLEMS Problem 6 - Solution A child pulls a sled up a snow-covered hill at a constant velocity with a force parallel to the hillside. If the sled has a mass of 9.5 kg, the hill slope is 12°, and the coefficient of friction between the sled and the snow is 0.20, what is the force applied by the child? y x FN Draw FBD F Hint 1: The normal force is always perpendicular to the surface. In this case, it is not opposite the force of gravity. Fk v θ = 12° Fgy Fg Fgx Hint 2: Choose directions as shown with the x in the direction of motion (up the ramp) and y in the direction of the normal. This greatly simplifies the math because we only have to break the force of gravity into components. The other forces all lie entirely along the x or y in the direction. We will need to find the components of the gravitational force (see the diagram to the right to verify that the angle between the y-direction and the force of gravity vector is 12°). y x θ = 12° Fgy mg cos Fgx mg sin θ = 12° Hint 3: Notice that the trig functions usually associated with the x and y- components are switched. This comes from the fact that the angle is measured relative to the y-axis instead of the x-axis. Fgy Fg Fgx Is the object accelerating? No, the child is pulling the sled at a constant velocity, so net force is 0. Define terms: m 9.5 kg, k 0.20, 12, Fnet 0, F ? We are interested in the force applied by the child so let’s look at the x-directions. Fnet , x F Fgx Fk 0 F Fgx Fk Creative Commons Attribution 3.0 Unported License 10 PHYSICS SEMESTER ONE UNIT 4: ANSWERS TO PROBLEMS We don’t know the friction force in this equation. This requires a look at the net force in the y-direction so we can find the normal force, and then the force of friction. Fnet , y FN Fgy 0 FN Fgy mg cos The friction is then Fk k FN k mg cos Insert this into the x-component force equation F Fgx Fk mg sin k mg cos 9.5 kg 9.8 m/s 2 sin12 0.20 9.5 kg 9.8 m/s 2 sin12 37.6 N The force from the child is 38 N up the hill. In this question, the actual calculations were left to the very end. This reduces the chance of having a round off error. In other problem solutions, one or two extra significant figures are kept until the final calculation. This is an old note but it doesn’t hurt to repeat it. Creative Commons Attribution 3.0 Unported License 11 PHYSICS SEMESTER ONE UNIT 4: ANSWERS TO PROBLEMS PROBLEM 7 Sleds A and B are connected by a horizontal rope, with A in front of B. Sled A is pulled forward with a horizontal rope with a tension of magnitude 29.0 N. The masses of sleds A and B are 6.7 kg and 5.6 kg, respectively. The magnitudes of kinetic friction on A and B are 9.0 N and 8.0 N respectively. B A TAB Tpull = 29.0 N i mB = 5.6 kg, FkB –8.0 N ˆi, mA 6.7 kg, FkA –9.0 N ˆi a) Find the acceleration of the two-sled system y x Draw FBD or FBDs choose directions FNA, FNB FkA, FkB Is the object accelerating? yes – need to find a Can we consider the sleds as one item? Sure. The sleds move together, with the same acceleration. There are cases where we would have to use two FBDs, like if we didn’t have the friction forces, but a combined FBD works in this case. Tpull v FgA, FgB The combined mass of the two sleds is m mA mB 5.6 kg 6.7 kg 12.3 kg Do we need to worry about the net force in the y-direction? In this case there is no movement in the y-direction (the net force in the y-direction is zero), and we don’t need to find the normal force to calculate the forces of friction. The y-direction forces have no components in the x-direction so we can then ignore them. The net force in the x-direction is the sum of the forces on the system. The directions of the friction forces are already taken into account in the variable definitions so they are added (added negative values instead of subtracted positive values.) Fnet , x Tpull FkA FkB max Creative Commons Attribution 3.0 Unported License 12 PHYSICS SEMESTER ONE UNIT 4: ANSWERS TO PROBLEMS Solving for the acceleration in the x-direction (the total acceleration because there is none in the ydirection). ax T pull FkA FkB m 29.0 N 9.0 N 8.0 N 12.3 kg 0.976 m/s 2 The sled is accelerating at 0.98 m/s2 in the positive x-direction. y x b) Find the tension on the rope connecting the sleds. Create the FBDs for each separate sled. The tension on the rope between the sleds doesn’t show up on the combined FBD. The magnitude of the tension is the same for each sled so we really on need one of the single sled FBDs. FNA Tpull FkA, TBA FgA v sled A Let’s look at sled B because it has fewer |forces. Once again, we ignore the y-direction forces so the forces are FkB and TAB . y The net force on sled B is x FNB Fnet , x TAB FkB mB ax TAB FkB Rearrange to solve for the tension TAB mB ax FkB 5.6 kg 0.976 m/s v 2 8.0 N FgB sled B 13.46 N The tension on the rope between the sleds is 13 N. We have done this by starting with two FBDs, one for each sled. This would have generated two equations with two unknowns a x and TAB . We could have then solved the two equations for the two unknowns. Creative Commons Attribution 3.0 Unported License 13 PHYSICS SEMESTER ONE UNIT 4: ANSWERS TO PROBLEMS A related question is one where two crates are pushing on each other (instead of pulling in this case). The force applied to the crates is also a push from behind the rear crate. The crates move at the same acceleration so the acceleration can be found by treating the crates as one (you may have to look at separate crates to find the friction forces). The force between the plates is Newton’s third law pair. It can be found by looking at one crate by itself. Another multiple crate problem has one crate stacked on the other. The force is from the side on one of the crates. The acceleration can be found from the combined FBD and the friction is for the bottom crate only. The friction force between the crates is a Newton’s third law pair. These questions usually ask for the maximum or minimum coefficient of friction between the crates. It can usually be found from the system acceleration and the FBD for the top crate. Creative Commons Attribution 3.0 Unported License 14 PHYSICS SEMESTER ONE UNIT 4: ANSWERS TO PROBLEMS PROBLEM 8 +y A loonie (mL = 6.99 g) and a dime (mD = 2.09 g) are attached to the ends of a thread. The thread lies over a smooth horizontal bar. Initially, the coins are held motionless. When released, the coins start to move. The friction between the thread and bar is negligible (zero). +y +y a) Find the acceleration of the coins. Draw FBDs TLD +y choose directions TDL +y FgD FgL Dime Loonie Note that the +y-direction is down for the loonie. This is consistent with the direction definition of the dime. Assume that the thread doesn’t stretch so downward motion of the loonie is the same as the upward motion of the dime. We only have motion and forces in the y-direction. The masses are defined in the question. From the tension properties TDL TLD or TLD TDL . This is not true if the thread goes over a movable wheel or if there is friction. Is the object accelerating? – yes, need to find a, use Newton’s second law. The forces on the dime are the gravitational force and the tension, FgD mD g and TLD . The net force on the dime is Fnet , D FgD TLD mD a All forces are in the same direction (±) so we can use scalars mD g TLD mD a Rearranging gives TLD mD a mD g Creative Commons Attribution 3.0 Unported License 15 PHYSICS SEMESTER ONE UNIT 4: ANSWERS TO PROBLEMS The forces on the loonie are FgL mL g and TDL TLD . The net force on the loonie is Fnet , L FgL TDL mLa mL g TLD mL a replace TDL with TLD mL g TLD mL a all vectors in same direction, replace with scalars Insert the expression for the tension TLD from the dime forces mL g mD a mD g mL a More rearranging gives mL a mD a mL g mD g a mL mD g mL mD Isolating the acceleration a g mL mD mL mD 9.80 m/s 6.99 2.09 2 6.99 2.09 5.289 m/s 2 The acceleration is 5.29 m/s2. b) Find the tension in the thread. We just sub our acceleration this back into one if the equations for the tension. First convert the dime mass to kg. mD = 2.09×10-3 kg TLD mD a mD g 2.09 103 kg 5.289 m/s 2 + 2.09 10 3 kg 9.80 m/s 2 3.15 102 N The tension on the tread is 3.1510-2 N. There are cases where you have more pulley ropes. There are cases where the rope goes over a fixed object or an object that moves. Check out examples in the assignments and text. Creative Commons Attribution 3.0 Unported License 16 PHYSICS SEMESTER ONE UNIT 4: ANSWERS TO PROBLEMS We could have treated the two masses as one, with the loonie weight as the force in one direction and the dime weight as the force in the other. That would have given us the acceleration. We would then use an individual coin FBD to find the tension in the tread. Can you apply Newton’s Laws to a fixed object? No, this only works for objects that can undergo acceleration. NANSLO Physics Core Units and Laboratory Experiments by the North American Network of Science Labs Online, a collaboration between WICHE, CCCS, and BCcampus is licensed under a Creative Commons Attribution 3.0 Unported License; based on a work at rwsl.nic.bc.ca. Funded by a grant from EDUCAUSE through the Next Generation Learning Challenges. Creative Commons Attribution 3.0 Unported License 17