Laws of Motion Compound Bodies Compound Bodies Two
Laws of Motion Compound Bodies
Compound Bodies Two or more masses experiencing the same motion. m 1 m 2 m 3 2 kg 4 kg 6 kg T = 24 N The three masses shown are attached to each other with strings and are being pulled to the right by another string exerting a tension of 24 N.
Forces Acting on All Objects Adding force vectors to all objects introduces complexity N 1 m 1 TA 2 kg w 1 N 3 N 2 TA m 2 TB 4 kg w 2 TB m 3 T = 24 N 6 kg w 3 However, the objects are all moving horizontally. The vertical forces acting on each object are opposite and equal, and cancel each other. As a result, we are only interested in the horizontal forces.
Internal and External Forces m 1 2 kg TA TA m 2 4 kg TB TB m 3 T = 24 N 6 kg Internal forces are the forces that act between the objects in a compound body. Here the internal forces are tension A in the string between mass 1 and mass 2, and tension B in the string between mass 2 and mass 3. External forces are forces that act on the entire compound body. In this case there is only one external force, the overall tension of 24 N pulling all three objects to the right.
Objects and systems m 1 2 kg TA TA m 2 4 kg TB TB m 3 T = 24 N 6 kg When working with a compound body, or a complex system, you can view and solve problems from two perspectives. Object: When working with a compound body you can treat each mass as a separate object and sum all the forces (internal and external) acting on each individual mass separately. The result will be several equations can then be solved by substituting the equations into each other (system of equations). System: An alternative approach involves treating the entire system as a single united object, and then summing only the external forces acting on the system as a whole. When this method is used, internal forces cancel. Some variables are best solved by summing forces for individual objects, while other variables are best solved by summing forces for the system.
Acceleration m 1 2 kg TA Acceleration can be solved using either strategy. TA m 2 TB 4 kg TB m 3 T = 24 N 6 kg Object Sum the forces for each object separately. When using this method internal forces DO NOT cancel. You now have three equations with three missing variables (system of equations). The equations must be substituted into one another.
Acceleration m 1 TA Acceleration can be solved using either strategy. TA 2 kg m 2 4 kg TB TB m 3 T = 24 N 6 kg Object Or, you can treat all the objects as one large aggregate object, or system. All object are moving together as one with the same acceleration. Now try summing the forces for the system.
Acceleration m 1 TA Acceleration can be solved using either strategy. TA 2 kg m 2 4 kg Object TB TB m 3 T = 24 N 6 kg System When solving for the system all the internal forces, tension A and tension B, are opposite and equal. All object are moving together as one with the same acceleration. When the forces for the system are summed internal forces will cancel and can be ignored.
Acceleration m 1 TA Acceleration can be solved using either strategy. m 2 TA 2 kg 4 kg Object TB TB m 3 T = 24 N 6 kg System All object are moving together as one with the same acceleration. Both methods work, but one is clearly easier.
The Net Force Acting on Each Mass m 1 2 kg TA TA m 2 TB 4 kg TB m 3 T = 24 N 6 kg Net force is synonymous with the sum of forces. While the net force is equal to all the force vectors acting on an object it is also equal to the objects mass multiplied by its acceleration.
Solving for the Internal Forces m 1 2 kg TA TA m 2 4 kg TB TB m 3 T = 24 N 6 kg Summing forces for the system causes the internal forces (tensions), to cancel, which would make it impossible to solve for tensions A and B. When solving for internal forces between objects, such as tension, you must sum the forces for each mass separately.
Pushed Masses Now the masses are being pushed by and external 24 N force. This time the internal forces will be normal forces due to the surfaces of the objects pressing against one another. F = 24 N m 1 m 2 m 3 2 kg 4 kg 6 kg
Acceleration F = 24 N msys = m 1 + m 2 + m 3 msys = 2 + 4 + 6 = 12 kg
Net Force Acting on Each Mass F = 24 N m 1 m 2 m 3 2 kg 4 kg 6 kg
Solving for the Internal Forces The masses all have surfaces pressing on one another, which create a pair of equal and opposite normal forces at each surface. When viewed as an entire system these normal forces cancel. F = 24 N m 1 m 2 m 3 2 kg 4 kg 6 kg N 2 on 1 N 1 on 2 N 3 on 2 N 2 on 3 The right surface of m 1 presses on m 2 with a normal force N 1 on 2 The left surface of m 2 presses back on m 1 with a normal force N 2 on 1 The right surface of m 2 presses on m 3 with a normal force N 2 on 3 The left surface of m 3 presses back on m 2 with a normal force N 3 on 2
Solving for the Internal Forces According to Newton’s Third Law, when two objects interact there is an opposite and equal force acting on each object. F = 24 N m 1 m 2 m 3 2 kg 4 kg 6 kg N 2 on 1 N 1 on 2 N 3 on 2 N 2 on 3 N 1 on 2 is equal and opposite N 2 on 1 N 2 on 3 is equal and opposite N 3 on 2 This means you do not need to solve all four normal forces. You only need to solve one normal force in each pair.
Solving for the Internal Forces When solving for the internal forces between masses in a compound body you must sum the forces acting on each mass separately. Then substitute the acceleration of the system that was solved earlier. F = 24 N m 1 m 2 m 3 2 kg 4 kg 6 kg N 2 on 1 N 1 on 2 N 3 on 2 N 2 on 3
Net Force on Each Mass Solve for the net force acting on each of the stacked masses. 2 kg The secret to this problem is asking what the objects are doing. What are they doing? Standing Still: 4 kg v=0 a=0 6 kg ΣF = 0 This is true for all the masses together and individually.
Normal Force Between Masses The blocks are not moving. Balanced forces. The Normal force at each surface has to equal the weight of the blocks above that surface. 2 kg 4 kg 6 kg
Normal Force Between Masses Solve for the net force acting on each of the hanging masses. The secret to this problem is asking what the objects are doing. 2 kg What are they doing? Standing Still: v=0 4 kg a=0 ΣF = 0 6 kg This is true for all the masses together and individually.
Tension in Strings Between Masses The blocks are not moving. Balanced forces. The tension in each string has to equal the weight of the blocks hanging from that string. 2 kg 4 kg 6 kg
- Slides: 21