5. (20 points) Newton\'s Laws Two bodies with nassless pulley. They slide along

Question

5. (20 points) Newton's Laws Two bodies with nassless pulley. They slide along inelinanswtirected yas sh ond the ad atber ty and acceleration of the body with mass mi eration and ve mass as 2 nected by a sn body with mass m2 in the direction of the body acee The coefficient of kinetic friction -0.1 betwee times less than the coefficient of kinetic friction plane. The angle -300 and the angle withh a are directed up., A forse icd to shown ocity the bo the y with -2 between 60° Find the magnitude of the force F. m, 0

Explanation / Answer

let T is the tension in the string that is connected between the blocks.

net force acting on m1,

Fnet1 = T - m1*g*sin(30) - N1*mue_k1

m1*a = T - m1*g*sin(30) - m1*g*cos(30)*mue_k1

==> T = m1*a + m1*g*sin(30) + m1*g*cos(30)*mue_k1

= 1*12 + 1*9.8*sin(30) + 1*9.8*cos(30)*0.1

= 17.7 N

now net force acting on m2,

Fnet2 = m2*g*sin(60) + F - N2*mue_k

m2*a = m2*g*sin(60) + F - m2*g*cos(60)*mue_k2

F = m2*a + m2*g*cos(60)*mue_k2 - m2*g*sin(60)

= 3*12 + 3*9.8*cos(60)*2*0.1 - 3*9.8*sin(60)

= 13.5 N <<<<<<<--------Answer

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