Question

1. on a building site, two buckets of paint of masses 10 kg each are attached to each other by a rope in the vertical as shown in the figure. find the force f and the tension in the rope between the buckets when

a) the buckets are at rest
b) the buckets are pulled upwards with an acceleration of 1 m/s²
(neglect the weight of the rope.)

Explanation:

Step1: Analyze the situation when buckets are at rest

When the buckets are at rest, the acceleration \(a = 0\).
Using Newton's second law \(F_{net}=ma\). For the two - bucket system (total mass \(M=10 + 10=20\space kg\)), \(F - Mg=Ma\). Since \(a = 0\), \(F=Mg\). Substituting \(M = 20\space kg\) and \(g = 9.8\space m/s^{2}\), \(F=20\times9.8=196\space N\).
For the lower bucket (mass \(m = 10\space kg\)), \(T - mg=ma\). Since \(a = 0\), \(T=mg\). Substituting \(m = 10\space kg\) and \(g = 9.8\space m/s^{2}\), \(T = 98\space N\).

Step2: Analyze the situation when buckets are pulled upwards with \(a=1\space m/s^{2}\)

For the two - bucket system (total mass \(M = 20\space kg\)), using \(F - Mg=Ma\).
\(F=M(g + a)\). Substituting \(M = 20\space kg\), \(g = 9.8\space m/s^{2}\) and \(a = 1\space m/s^{2}\), \(F=20\times(9.8 + 1)=20\times10.8=216\space N\).
For the lower bucket (mass \(m = 10\space kg\)), using \(T - mg=ma\).
\(T=m(g + a)\). Substituting \(m = 10\space kg\), \(g = 9.8\space m/s^{2}\) and \(a = 1\space m/s^{2}\), \(T=10\times(9.8+ 1)=108\space N\).

Answer:

a) When the buckets are at rest, \(F = 196\space N\) and \(T = 98\space N\).
b) When the buckets are pulled upwards with \(a = 1\space m/s^{2}\), \(F = 216\space N\) and \(T=108\space N\).