Three masses, m1 = 10 kg, m2 = 8 kg, and m3 = 6 kg, are connected as shown in Fig. 7-16. Neglecting friction, what is the acceleration of the system? What are the tensions in the cord on the left and in the cord on the right? Would the acceleration be the same if the middle mass m2 were removed?
Total mass of system = (10 + 8 +6) = 24 kg
Resultant Force on system = m1g – m3g
The normal force N balances m2g; SF = mTa
m1g – m3g = (m1 + m2 +m3)a ; (10 kg)(9.8 m/s2) – (6 kg)(9.8 m/s2) = (24 kg) a
(24 kg)a = 98.0 N – 58.8 N; a = 1.63 m/s2 ; The acceleration is not affected by m2.
To find TA apply SF = m1a to 10-kg mass: m1g – TA = m1a ; TA = m1g – m1a
TA = m1(g – a) = (10 kg)(9.8 m/s2 - 1.63 m/s2); TA = 81.7 N
Now apply to 6-kg mass: TB – m3g = m3a; TB = m3g + m3a
TB = (6 kg)(9.8 m/s2 + 1.63 m/s2) ; TB = 68.6 N
fisica con ordenador
Total mass of system = (10 + 8 +6) = 24 kg
Resultant Force on system = m1g – m3g
The normal force N balances m2g; SF = mTa
m1g – m3g = (m1 + m2 +m3)a ; (10 kg)(9.8 m/s2) – (6 kg)(9.8 m/s2) = (24 kg) a
(24 kg)a = 98.0 N – 58.8 N; a = 1.63 m/s2 ; The acceleration is not affected by m2.
To find TA apply SF = m1a to 10-kg mass: m1g – TA = m1a ; TA = m1g – m1a
TA = m1(g – a) = (10 kg)(9.8 m/s2 - 1.63 m/s2); TA = 81.7 N
Now apply to 6-kg mass: TB – m3g = m3a; TB = m3g + m3a
TB = (6 kg)(9.8 m/s2 + 1.63 m/s2) ; TB = 68.6 N
fisica con ordenador
