Question

Consider the scenario shown in the diagram below. A box (mass m1) is on a slope, which is inclined at an angle theta with respect to the horizontal. The box is connected via a string and pulley to a hanging weight (mass m2). The string and the pulley are assumed to be massless and frictionless, but there is friction between the box and the incline (with coefficient of kinetic friction mu k). Find an expression for the tension T in the string as the box slides down the incline Your answer should be written in terms of m1, m2, theta, muk, and g.T =

Answer 1

Let T is the tension in the string.

lt a is the accleration of the blocks.

net force acting acting m2, Fnet2 = m2*g - T

m2*a = m2*g - T (using Netwon's second law)

a = g - T/m2   ---(1)

net force acting on m1, Fnet1 = T + m1*g*sin(theta) - mue_k*N

m1*a = T + m1*g*sin(theta) - mue_k*m1*g*cos(theta)

a = T/m1 + g*sin(theta) - mue_k*g*cos(theta)   ---(2)

from equations 1 and 2

T/m1 + g*sin(theta) - mue_k*g*cos(theta) = g - T/m2

T/m1 + T/m2 = g - g*sin(theta) + mue_k*g*cos(theta)

T*(m1+m2)/(m1*m2) = g - g*sin(theta) + mue_k*g*cos(theta)

T = (m1*m2)*(g - g*sin(theta) + mue_k*g*cos(theta))/(m1+m2) <<<<<--------Answer