Three identical blocks connected by ideal strings are being pulled along a horiz

Question

Three identical blocks connected by ideal strings are being pulled along a horizontal frictionless surface by a horizontal force F_vec The magnitude of the tension in the string between blocks B and C is T = 3.00{ m N} . Assume that each block has mass m = 0.400{ m kg} .What is the magnitude F of the force?What is the tension T_AB in the string between block A and block B?

Explanation / Answer

The answer depends on the arrangement of the blocks. Assuming they're in order, it could be: [A]---[B]---[C]------>f or [C]---[B]---[A]------>f A primer on ideal strings: They do not stretch They have no mass they transmit force directly between their endpoints (The left end pulls right with a force equal to the tension, and the right end pulls left with a force equal to the tension.) The answers will be different depending on the two possible cases. I'll assume that it's the first case ([A]---[B]---[C]------>f) and let you work it out if it's different. Because the strings do not stretch, all three blocks will have the same acceleration. Because the blocks are all identical and the accelerations are the same, the resultant force on each block must be identical ((Fa = Fb = Fc). From that, we can determine the force on each string and the total force. Assuming [A]--Tab--[B]--Tbc--[C]---f--->: it is given that tension between B and C (Tbc) is 3. (I'll assume that is in Newtons). We therefore have: [A]--Tab--[B]--3--[C]---f---> If we use the convention of + is right and - is left, we can say: Fb = (+3) + (-Tab) Fa = Tab Fb = Fa so: (+3) + (-Tab) = Tab Solve for Tab (+3) = Tab - (-Tab) (+3) = 2*Tab Tab = 1.5N That leaves us knowing the following: [A]--1.5--[B]--3--[C]---f---> We can now find the resultant force on any of the blocks: Fa = Tab Fa = 1.5N since Fa = Fb = Fc, we can also say that: Fb = 1.5N Fc = 1.5N since we know that Tbc = 3N, we can solve for F Fc = f + (-Tbc) 1.5N = f + (-3 N) 1.5N - (-3 N) = f f = 4.5N Unless you also need to solve for the acceleration of the system in another part of the question, the listed mass of the block is extraneous information. Just in case you need that, we can say: f = mA 4.5N = (3 * 0.4kg) * A (1 N = 1 kg*m/s^2) (4.5 kg*m/s^2) / (1.2 kg) = A A = 3.75 m/s^2

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