Answer to Question #109078 in Classical Mechanics for Steve Clayton

Question

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?

Answer 1

$M_{1}\times a=T-F_{fr1}-M_{1}\times g\times\sin\alpha;N_1=M_{1}\times g\times\\cos\alpha;\\M_{2}\times a=T-F_{fr2}-M_{2}\times g\times\sin\beta;N_2=M_{2}\times g\times\cos\beta;\\T-\mu_{s}\times M_{1}\times g\times\cos\alpha-M_{1}\times g\times\sin\alpha=0;\\-T+\mu_{s}\times M_{2}\times g\times\cos\beta+M_{2}\times g\times\sin\beta=0;\\\mu_{s}\times M_{2}\times g\times\cos\beta-\mu_{s}\times M_{1}\times g\times\cos\alpha+M_{2}\times g\times\sin\beta-M_{1}\times g\times\sin\alpha=0\\\mu_{s}\times( M_{2}\times g\times\cos\beta-\ M_{1}\times g\times\cos\alpha)+g\times (M_{2}\times \sin\beta-M_{1}\times \sin\alpha)=0\\\mu_{s}=\frac{M_{1}\times \sin\alpha-M_{2}\times \sin\beta}{M_{1}\times cos\beta-M_{2}\times \cos\alpha}=\frac{4.5\times 0.86-6.4\times 0.44}{6.4\times 0.9-4.5\times 0.51}=0.3;\mu_{s}>0.129$

Anser: the the system is static