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Force Systems. Combination Systems – connected masses Horizontal Pulley Atwood’s Machine. For any force system you must sum forces. F net = S F = F 1 + F 2 … ma = F 1 + F 2 …. Hwk Sheet: Problems in Force 2. prb 4 – 7. Connected Masses.
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Force Systems • Combination Systems – connected masses • Horizontal Pulley • Atwood’s Machine
For any force system you must sum forces. Fnet = SF = F1 + F2 … ma = F1 + F2 …
Fnet = mtotaa = Fnet. m1 + m2 + m3 . A constant net force, F, must accelerate the entire system’s mass.
Sketch free body diagrams for each mass ignore vertical forces. Assign 1 direction as positive (right). Write the Fnet equation for each, find acceleration.then isolate each masses to find T1 & T2. m1a = T1. m2a = T2 - T1. m3a = F – T2.
Ex 1: Connected Masses: Given a Fnet of 20N and masses of 4, 3, and 1 kg, find the acceleration of the system and the tension in each cord. a = Fnet. m1 + m2 + m3
Find system acceleration: a = Fnet m1 + m2 + m3. 20 N (4 + 3 + 1) kga = 2.5 m/s2.
Use the free body diagram & known acceleration to find the tension in each cord. 4 kg T1 = m1a = (4 kg)(2.5 m/s2) = 10 N. F -T2 = m3a or F - m3a = T2. 20N - (1 kg)(2.5 m/s2) = 17.5 N 1 kg
Check the calculation using the 3rd mass.T2 – T1 = m2a 17.5 N – 10 N = 7.5 Nm2a = (3 kg)(2.5 m/s2) = 7.5 N.It is correct!!
m1a = T1. m2a = T2 - T1. • T1 = 10 N • T2 = 17.5 N
The masses accl together, the tension is uniform, accl direction is positive.
-T. M2. m2g Sketch free body diagrams for each mass separately. Write Newton’s 2nd Law equation for each. Fn. +T. M1. m1g m1a = T m2a = m2g - T
Add the equations: m1a + m2a = T + m2g – T T cancels. m1a + m2a = m2g Factor a & solve
a = m2g m1 + m2 Solve for a, and use the acceleration to solve for the tension pulling one of the masses. m1a = T
Ex 2: Horizontal Pulley: Given a mass of 4 kg on a horizontal frictionless surface attached to a mass of 3 kg hanging vertically, calculate the acceleration, and the tension in the cord.Compare the tension to the weight of the hanging mass, are they the same?
a = 4.2 m/s2T = 16.8 Nmg = 30 N, it is less than the tension.
To practice problems go to:Hyperphysics site.Click Mechanics, Newton’s Laws, Standard problems, then the appropriate symbol. http://hyperphysics.phy-astr.gsu.edu/hbase/hph.html#mechcon
Atwood’s Machine Use wksht.
Given Atwood’s machine, m1 = 2 kg, m2 = 4 kg. Find the acceleration and tension.
Using the previously determined accl, the force F2 acting on the smaller mass is F2 = m2a
By Newton’s 3rd Law, F2 acts backward on m1.The force on m1 is: The net force, F1, on m1 is: m1 F2 F
Given a force of 10N applied to 2 masses, m1 =5 kg and m2 =3kg, find the accl and find F2 (the contact force) between the boxes. a = 1.25m/s2 F2 = 3.75 N
Given a force of 100 N on 100 1 kg boxes, what is the force between the 60th and 61st box. 100-N 1-kg
Find a for system.F2 must push the remaining 40 boxes or 40 kg. 40 N.
Ignoring friction, derive an equation to solve for a and T for this system: Begin by sketching the free body diagram Write the equations for each box Add them. Solve for accl
Given a 30o angle, and 2 masses each 5-kg,find the acceleration of the system, and the tension in the cord. a= 2.45m/s2. T =36.75 N
Derive an equation for the same inclined pulley system including friction.