Find the tension on the string?

In this case you're going to want to do a sum of forces in the x direction on mass one, and then whatever's left over is what's pulling on mass 2, since the friction will take away from the force pulling to the right on mass one. Once you have the force that's acting on the wire to the upper right on mass two, you can add the friction caused by m2's weight. The sum of that friction + the force exerted by the pulling force will be the tension in the rope.

Make sure you're drawing a free body diagram for all the forces acting on each mass. Its easier to think about if you solve for the sum of forces acting on m1 first, and then solve the sum of forces acting on m2.

T*sin theta = m*(w^2)*r -------------------------- a million and T*cos theta = m*g --------------------------------- 2 the place theta = semi-vertical attitude of the cone which the string describes = tan^-a million[30/40] = tan^-a million[3/4] and w is the angular speed of the around action of the conical pendulum Dividing a million by technique of two, we get tan theta = 3/4 = [(w^2)*r]/g or w^2 = 3*g/(4*r). So a million could be rewritten as T*sin theta = m*(w^2)*r= (3*m*g)/4 Squaring and including a million and 2 we get T^2 = [(mg)^2]*[(3/4)^2 + a million^2] or T = 0.a million*9.8sq rt[9/sixteen +a million] = (5*0.a million*9.8)/4 = a million.225 N