2nd Law - Relationship between F, m, a
Unbalanced forces
ΣF_{net} = ma F_{net} = Resultant
Ex) Mass M below is lifted up with an acceleration a. Find the tension in the string T ΣF_{net} = ma
T - mg = ma
T = ma + mg T = m(g + a)
T = m(g - a) Two masses are connected by a string. Find tension force on top and bottom rope.
Ex 1) Two masses are connected by a string and remain at rest. Find tension force on top and bottom rope. SF = 0 Top String Tension Force
T_{A} - m_{a}g - m_{b}g = 0
T_{A} = m_{a}g + m_{b}g
Bottom String Tension Force
T_{B} - m_{b}g = 0 T_{B} = m_{B}g Ex 2) Find the tension of top and bottom string if objects accelerated upward
Free body diagram B (easier) Sketch all forces on A Force of tension on B (easier) Free body diagram B Sketch all forces on B
SF = m_{B}a T_{B} - m_{B}g = m_{B}a
T_{B} = m_{B}a + m_{B}g = (force of accel. of B + weight of B)
Sketch all forces on A
SF = m_{A}a T_{A} - T_{B} - m_{A}g = m_{A}a
T_{A} = m_{A}a + m_{A}g + T_{B}
Since T_{B} = m_{B}a + m_{B}g T_{A} = m_{A}a + m_{A}g +m_{B}a+m_{B}g)
T_{A}=(m_{A}a+m_{B}a)+(m_{B}g+m_{A}g) = (force of accel) + (weights)
Giancoli p. 98) 8, 25
Choice 2
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