Bob has three bags connected by two ropes of equal strength. He pulls the bags on a rough horizontal ground towards the left, as shown. Which rope will break first?
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But if bag 3 weighs 1 ton and bag 2 is empty, the difference is negligible and either rope has an equal chance of breaking first.
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Even if the difference will be negligible, still rope 2 will have slightly smaller tension in comparison to rope 1, unless you assume bag 2 as massless. There is no point to assume bag 2 as massless as that will suggest that bag 2 does not exist and that will change the problem.
How about if the masses are so small that the friction becomes more or less negligible? Then none of the rope will break. You cannot have friction without reaction force of weight from ground. You tried to make a vertical question horizontal which is fine, but you have to set the conditions right, particularly when you have an answer saying it depends on mass of the bags.
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Even if the friction is neglected still the rope 1 will break first. In this case T 1 = ( m 2 + m 3 ) a while T 2 = m 3 a and hence T 1 is still greater than T 2 .
Rope one will break first because it has the strain of bob pulling on it, while rope two only has a rope pulling on it
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Let the bags move by an acceleration a and f 1 , f 2 , f 3 are the friction forces that acts on the bags as shown in the diagram. Let the tensions in the rope 1 and 2 be T 1 and T 2 respectively. Let the masses of the bags 1,2 and 3 be m 1 , m 2 and m 3 respectively.
Now, the rope 1 pulls the bags 2 and 3, therefore, applying the Newton's second law , we can write, T 1 = f 2 + f 3 + ( m 2 + m 3 ) a .
The rope 2 pulls only bag 3, therefore, T 2 = f 3 + m 3 a .
Clearly, T 1 > T 2 , therefore, the rope 1 will break first .