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accelerations and tension question PHYSICS?
I'm struggling with this question:
Suppose two boxes on a frictionless table are connected by a heavy cord of mass 1.0 kg. Calculate the acceleration (magnitude) of each box and the tension (magnitude) at each end of the cored, using the free-body diagrams shown in Figure 4-53. Assume FP = 42.0 N, ma = 9.0 kg, and mb = 13.5 kg, and ignore sagging of the cord. Compare your results to Example 4-12 and Figure 4-22.
find:
mA acceleration in m/s2
mB acceleration in m/s 2
FTA in Newtons
FTB in Newtons
fig 4-53
http://www.webassign.net/giancoli/4-53alt.gif
fig 4-22
4 Answers
- Anonymous9 years agoFavorite Answer
FBT - FTB + FTA - FAT +FP = 0
13.5a -1a +1a -9a +42 = 0
a = 9.33
mA = 9.33
mB = 9.33
FTA = 9.33 x 9 -13.5 x 9.33 = -41.985
FTB = 13.5 x 9.33 - 9.33 x 9 = 41.985 [otherwise the cord would sag]
fig 4-22 represents a similar situation, I guess!
- Anonymous4 years ago
evaluate each mass in my opinion and write equations for each from Newton's 2nd regulation: m?a = m?g - T (a million): the place a is the acceleration of the block and T is the rigidity in the twine. Now for m?: m?a = T - m?g (2) you should undergo in innovations that on the left acceleration is up, yet on the phenomenal component it is down. This motives rigidity to artwork in opposite guidelines on the two component. including the two equations at the same time: m?a+m?a = m?g - m?g (the T's cancle) a(m?+ m?) = g(m?- m?) a = g(m?- m?) / (m?+ m?) Plugging in unknowns supplies: a = -9.8m/s² (5.0kg - 8.0kg) / (5.0kg + 8.0kg) = 2.3m/s² Now, you may basically plug in this fee in between the 1st 2 equations, (a million) or (2), above and resolve for T. you're able to do this area!
- Anonymous4 years ago
evaluate each mass in my view and write equations for each from Newton's 2d regulation: m?a = m?g - T (a million): the place a is the acceleration of the block and T is the rigidity interior the twine. Now for m?: m?a = T - m?g (2) you're able to shop in mind that on the left acceleration is up, yet on the excellent component that's down. This motives rigidity to artwork in opposite guidelines on the two component. including the two equations jointly: m?a+m?a = m?g - m?g (the T's cancle) a(m?+ m?) = g(m?- m?) a = g(m?- m?) / (m?+ m?) Plugging in unknowns provides: a = -9.8m/s² (5.0kg - 8.0kg) / (5.0kg + 8.0kg) = 2.3m/s² Now, you could purely plug in this fee in between the 1st 2 equations, (a million) or (2), above and resolve for T. you're able to try this area!
