Applying Newton's Laws — Question 4

Question 4

Three connected blocks reveal how tension transmits acceleration

Three blocks of masses 2, 3 and 5 kg are linked in that left-to-right order by light taut horizontal cords on a frictionless table. A 30 N force pulls the 5 kg block rightward.

1. Find the common acceleration.

2. Find each cord tension.

3. If the 30 N pull were instead leftward on the 2 kg block, find the new tension magnitudes.

Question 4 - Solution

Three connected blocks reveal how tension transmits acceleration

(a) Use the complete system.

$$30=(2+3+5)a\ \Rightarrow\ \boxed{a=3\,\mathrm{m/s^2}}$$ Total mass is 10 kg, so the acceleration magnitude is 3 m/s$^2$.

(b) Isolate blocks left of each connector.

$$T_{23}=2a=6\,\mathrm N,\quad T_{35}=(2+3)a=15\,\mathrm N$$ The cord between 2 and 3 kg pulls only the 2 kg subsystem, so $T_{23}=2(3)=6$ N. The cord between 3 and 5 kg pulls the 5 kg combined left subsystem, so $T_{35}=(2+3)3=15$ N.

(c) Reverse which mass each cord must pull.

$$T'_{35}=5|a|=15\,\mathrm N,\quad T'_{23}=(3+5)|a|=24\,\mathrm N$$ With the force on the leftmost block, acceleration is 3 m/s$^2$ leftward. The right connector pulls the 5 kg block, so its tension is 15 N. The left connector pulls the 3+5 kg subsystem, so its tension is 24 N. Same total acceleration magnitude does not fix every internal force.