Question #109078
This web site https://www.geogebra.org/m/ZeZjAfta is for double incline plain calculations. It list a rope tension (t = 22.5 s). I need to know how much force in poundage that is? I have a screen shot of the calculation. I see no ability for an attachment. I this example the tension is increasing as one weight pull the other down the incline?
1
Expert's answer
2020-04-20T10:13:56-0400

M1×a=TFfr1M1×g×sinα;N1=M1×g×cosα;M2×a=TFfr2M2×g×sinβ;N2=M2×g×cosβ;Tμs×M1×g×cosαM1×g×sinα=0;T+μs×M2×g×cosβ+M2×g×sinβ=0;μs×M2×g×cosβμs×M1×g×cosα+M2×g×sinβM1×g×sinα=0μs×(M2×g×cosβ M1×g×cosα)+g×(M2×sinβM1×sinα)=0μs=M1×sinαM2×sinβM1×cosβM2×cosα=4.5×0.866.4×0.446.4×0.94.5×0.51=0.3;μs>0.129M_{1}\times a=T-F_{fr1}-M_{1}\times g\times\sin\alpha;N_1=M_{1}\times g\times\\cos\alpha;\\M_{2}\times a=T-F_{fr2}-M_{2}\times g\times\sin\beta;N_2=M_{2}\times g\times\cos\beta;\\T-\mu_{s}\times M_{1}\times g\times\cos\alpha-M_{1}\times g\times\sin\alpha=0;\\-T+\mu_{s}\times M_{2}\times g\times\cos\beta+M_{2}\times g\times\sin\beta=0;\\\mu_{s}\times M_{2}\times g\times\cos\beta-\mu_{s}\times M_{1}\times g\times\cos\alpha+M_{2}\times g\times\sin\beta-M_{1}\times g\times\sin\alpha=0\\\mu_{s}\times( M_{2}\times g\times\cos\beta-\ M_{1}\times g\times\cos\alpha)+g\times (M_{2}\times \sin\beta-M_{1}\times \sin\alpha)=0\\\mu_{s}=\frac{M_{1}\times \sin\alpha-M_{2}\times \sin\beta}{M_{1}\times cos\beta-M_{2}\times \cos\alpha}=\frac{4.5\times 0.86-6.4\times 0.44}{6.4\times 0.9-4.5\times 0.51}=0.3;\mu_{s}>0.129


Anser: the the system is static

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