My question, Find the magnitude of the contact force between boxes 1 and 2,betwe

Question

My question, Find the magnitude of the contact force between boxes 1 and 2,between boxes 2 and 3. My answer, For boxes 1 and 2. I know that acceleration should be the same between theboxes. acceleration= N / total mass.                      7.50 / 1.30 +3.20 = 7.50 N / 4.5 kg acceleration = 1.66 = 1.7 m/s2 mass of box 2 = 3.20 kg x acceleration = 1.7m/s2 3.20 x 1.7 = 5.44 N I assume perform the same calculation to find the forcebetween boxes 2 and 3, however, I do not think that my work andanswer is 100% correct, could someone please show me where I amgoing wrong. force of magnitude 7.50 N pushes three boxes with masses m1 = 1.30 kg , m2 = 3.20 kg , and m3 = 4.90 kg . Find the magnitude of the contact force between boxes 1 and 2,between boxes 2 and 3. My answer, For boxes 1 and 2. I know that acceleration should be the same between theboxes. acceleration= N / total mass. 7.50 / 1.30 +3.20 = 7.50 N / 4.5 kg acceleration = 1.66 = 1.7 m/s^2 mass of box 2 = 3.20 kg x acceleration = 1.7m/s^2 3.20 x 1.7 = 5.44 N I assume perform the same calculation to find the forcebetween boxes 2 and 3, however, I do not think that my work andanswer is 100% correct, could someone please show me where I amgoing wrong.

Explanation / Answer

M = 1.3 + 3.2 + 4.9 = 9.4 kg a = 7.5 / 9.4 = .80 m/s2   for all 3boxes F12 = (3.2 + 4.9) * .8 = 6.48N     force required to accelerate boxes 2& 3 F23 = 4.9 * .8 = 3.92 N    force required to accelerate box 3 Check: Net force on box 2 is F2 = 6.48 - 3.92 = 2.56N a = 2.56 / 3.2 = .8 m/s2 You can also do the same for box 1 You can also do the same for box 1

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