I have 3 different blocks. All connected by a string. I haveone that is laying o

Question

I have 3 different blocks. All connected by a string. I haveone that is laying on a horizontal surface, one is on an inline,and one is hanging in the air. I know the following M1=6kg M2=8kg M3=4Kg Coeffcient of friction = 0.2 Fa=110 N at 60 degrees Angle of incline= 30 degrees
I have to answer
1. What direction is the system moving 2. What is the friction force acting on M1 and M2 3. What is the acceleration of each mass 4. What is the tension in each rope? 5. What is the net force on the system? 6. What is the net force on each mass?

Explanation / Answer

the force acting on block 1 are y direction Fy = n1 + (-w1) = 0 or n1 = w1 = m1 * g ------------(1) x direction Fx = T + (-Fax - fk) = T - (Fa * cos -k* n1) --------------(2) from (1) and (2) we get Fx = T - (Fa * cos - k* m1 *g) m1 * ax = T - (Fa * cos -k* m1 * g) or ax= (1/m1) * (T - (Fa * cos -k* m1 * g)) ---------------(3) = 60o the forces acting on block 2 are y direction Fy = n2 + (-w2) = 0 or n2 = w2 = m2 * g ------------(4) x direction Fx = w1 - fk = w1 - k*n2 here,Fx = T or T = w * sin1 - k* n2 from (4) and (5) we get T = w * sin1 - k* m2 * g = m1 * g* sin1 - k* m2 * g --------------(5) the forces acting on block 3 are y direction Fy = n3 + (-w3) = 0 or n3 = w3 = m3 * g -------------(6) here,n3 = T1 or T1 = m3 * g ---------------(7) 1.the tension acting on mass m3 is more than the tension inthe upper two ropes therefore,the system is moving in the downwarddirection. 2.the friction force acting on M1 and M2are fk= k* m1 * g and fk= k* m2 * g 3.from equations (3) and (5) we get ax= (1/m1) * (m1 * g * sin1 -k* m2 * g - (Fa * cos - k*m1 * g)) 1 = 30o the above equation gives the acceleration of each mass 4.the tension in each of the two upper ropes is T = m1 * g * sin1 - k* m2 * g 5.the net force on the system is Fnet = T1 - T 6.the net force on each mass is F = m1 * ax = (T - (Fa * cos -k* m1 * g)) or F = (m1 * g * sin1 - k* m2 * g - (Fa* cos - k* m1 * g)) the forces acting on block 2 are y direction Fy = n2 + (-w2) = 0 or n2 = w2 = m2 * g ------------(4) x direction Fx = w1 - fk = w1 - k*n2 here,Fx = T or T = w * sin1 - k* n2 from (4) and (5) we get T = w * sin1 - k* m2 * g = m1 * g* sin1 - k* m2 * g --------------(5) the forces acting on block 3 are y direction Fy = n3 + (-w3) = 0 or n3 = w3 = m3 * g -------------(6) here,n3 = T1 or T1 = m3 * g ---------------(7) 1.the tension acting on mass m3 is more than the tension inthe upper two ropes therefore,the system is moving in the downwarddirection. 2.the friction force acting on M1 and M2are fk= k* m1 * g and fk= k* m2 * g 3.from equations (3) and (5) we get ax= (1/m1) * (m1 * g * sin1 -k* m2 * g - (Fa * cos - k*m1 * g)) 1 = 30o the above equation gives the acceleration of each mass 4.the tension in each of the two upper ropes is T = m1 * g * sin1 - k* m2 * g 5.the net force on the system is Fnet = T1 - T 6.the net force on each mass is F = m1 * ax = (T - (Fa * cos -k* m1 * g)) or F = (m1 * g * sin1 - k* m2 * g - (Fa* cos - k* m1 * g))

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