Three blocks on a frictionless surface are connected by massless strings, with M1 = 1.60kg, M2 = 2.10kg, and M3 = 3.70kg. Due to the force F acting on M3, the system accelerates to the right. Given that T1 is 4.10Newtons, calculate T2.
M1-----(T1)-----M2-----(T2)-------M3
That's the best picture I can do.
Anywho, I would assume that the tension would be equal but my answers are being rejected so I don't really know what to do
Thanks in advance:)
M1-----(T1)-----M2-----(T2)-------M3
That's the best picture I can do.
Anywho, I would assume that the tension would be equal but my answers are being rejected so I don't really know what to do
Thanks in advance:)
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F - T2 = M3*a
T2 - T1 = M2*a
F - T1 = (M2+M3)a
F - T1 = (M2+M3)/(M1+M2+M3) * F
M1/(M1+M2+M3) * F = T1
F = (M1+M2+M3)/M1 * T1
F = 7.4/1.6 * 4.10 N = 18.96 N
T2 = F - M3*a
T2 = F - M3/(M1+M2+M3) * F
T2 = (M1+M2)/(M1+M2+M3) * F
T2 = 3.7/7.4 * 18.96 N = 9.48 N
T2 - T1 = M2*a
F - T1 = (M2+M3)a
F - T1 = (M2+M3)/(M1+M2+M3) * F
M1/(M1+M2+M3) * F = T1
F = (M1+M2+M3)/M1 * T1
F = 7.4/1.6 * 4.10 N = 18.96 N
T2 = F - M3*a
T2 = F - M3/(M1+M2+M3) * F
T2 = (M1+M2)/(M1+M2+M3) * F
T2 = 3.7/7.4 * 18.96 N = 9.48 N