I am given this and the fact that acceleration = 2.00 m/s squared, and we are to
ID: 1694975 • Letter: I
Question
I am given this and the fact that acceleration = 2.00 m/s squared, and we are to assume that the coefficients of kinetic friction, µ, between block and incline are same for both inclines. So I calculated tension and got 28.6 N, which my teacher says is wrong. She won't show me the solution or the answer bc she has no time to. Can someone please show me step by step solution on how to correctly do it? Here is what I did (and is wrong, but I don't know why.): I separated the triangle into two components. Focusing on the 9kg mass, I put Fparallel is facing down while F Friction and FTension opposes it, so (FII - Ff - FT = ma)FII is just mgsin(theta), Ff is just µmgsin(theta), and FT is what we're solving for...oh, and ma is just the whole block accelerating. I also attempted to find the coefficient of friction, and got 0.15 which I'm fairly certain is right. mgsin(theta) - µmgsin(theta) - FT = ma (9)(9.8)(sin 40) - (0.15)(9)(9.8)(cos40) - FT = (9)(2) after all the math I got 28.6 N for FT.
Explanation / Answer
Taking the assumption , that 9Kg is sliding downwards and 4Kg is sliding upwards/ Net force experienced by 9Kg mass> F = mgSin40 - T-µmgCos40 { where tension in the rop} Putting the values give, 9(2) = 9(9.8)(0.642) -T - µ(9)(9.8)(0.766) 18 = 56.6244 -T - µ67.56 T = 38.6244 -µ (67.56)----------------eq(1) Net force experienced by 4Kg mass, F = T -mgSin 40 - µmg cos40 Putting the values give, 4(2)= T - 4(9.8)(0.642) - µ(4)(9.8)(0.766)---------eq(2) 8= T- 25.1664 - µ30 T = 33.1664 + µ30 ------------eq(3) Putting the value of (T) from eq(1) in eq(3) gives, 38.6244 -µ (67.56) = 33.1664 + µ30 5.458 =µ ( 30 + 67.56) 5.458/97.56=µ µ= 0.0559 approx Putting the value of coefficient of friction in eq(1), gives T= 38.6244 -(0.0559) (67.56) T= 38.6244 - 3.7796 T= 34.8448 N approxRelated Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.