Force on Blocks: Fig. 5-1a & b, Mass 30 kg, Forces 28 & 14 N

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In the discussion about the forces acting on two blocks with a total mass of 30 kg, Fig. 5-1a illustrates a scenario where a constant horizontal force Fa is applied to block A, resulting in a 28.0 N force exerted on block B. Conversely, Fig. 5-1b depicts block B receiving the same force Fa, leading to a 14.0 N force exerted on block A. To find the acceleration of the blocks in Fig. 5-1a, the equation F=ma is used, yielding an acceleration of 28 N divided by 30 kg. It is emphasized that the same force Fa acts on both configurations, affecting the total mass equally. Understanding these dynamics is crucial for solving the forces and accelerations involved.
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Fig. 5-1a shows two blocks on a smooth surface. A constant horizontal force Fa is applied to block A, which pushes against block B with a 28.0 N force horizontally to the right. In Fig. 5-1b, the same force Fa is applied to block B so that block A pushes on block B with a 14.0 N force horizontally to the left. The blocks have a total mass of 30.0 kg.

W0074-N.jpg


Fig. 5-1

(a) What is the magnitude of the acceleration of the blocks in Fig. 5-1a?

what i tried was F=ma so 28=30a so 28/30=a

(b) What is the magnitude of the force Fa?
 
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B-80 said:
(a) What is the magnitude of the acceleration of the blocks in Fig. 5-1a?

what i tried was F=ma so 28=30a so 28/30=a
28.0 N is the force on block B (in case a), not on both blocks. You have to use all the information given to solve this. Hint: The same force pushes the same total mass in each case.
 
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