If you treat the whole system as one solid thing, you will get the answer.
FP= mtotal x acceleration
44= (11+13.5+1)a
44=25.5a
a=1.73m/s^2
Use this acceleration value to find FTA and FTB using Newton's 2nd Law.
FBT=13 x 1.73
FBT= 22.5 N
(FBT and FTB are equal in magnitude)
and
44-FAT=11 x 1.73
44-FAT=19.03
FAT=44-19.03
FAT=25.0 N
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 = 44.0 N, ma = 11.0 kg, and mb = 13.5 kg, and ignore sagging of the cord. Compare your results to Example 4-12 and Figure 4-22.
mA acceleration ____ m/s2
mB acceleration ____ m/s2
FTA ____ N
FTB ____ N
1 answer
1 answer