Question

I really need help with this. Anyone?
-Attached.

One end of a light inextensible string is attached to a fixed point A of a fixed vertical wire. The other end ins attached to a small ring B, of mass 0.2 kg, through which the wire passes. A horizontal force of magnitude 5N is applied to the mid-point M of the string. The system is in equilibrium with the string taut, with B below A and with angles BAM and ABM equal to 30 degrees.

Answer 1

a) Show that tension in BM is 5N.
b) The ring is on the point of sliding up the wire. Find the coefficient of friction between the ring and the wire.

Answer 2

can u draw forces at M?

Answer 3

take the M point and draw there the forces...you have 3 different forces...one is the horizontal force of 5N and 2 are the tensions...one at AM and one at BM...both pointing upwards

Answer 4

then you need to take the Σχ=0 and the balance of forces on the x-plane.2 tips there=the tension exerted at BM is the same as that exerted on AM for the magnitude.

Answer 5

so it would be something like T1cosθ + T2cosθ=5Ν, where θ=180-90-30(from the 2 triangles), then as i said because T1=T2 then the tension would be 5N

Answer 6

@electrokid : Can you help me with the second part?

Answer 7

I dont get it. What force would make the ring go up?

Answer 8

I think the tension in the string BM has to be resolved.

Answer 9

T cos 30 would make the ring go up, I guess. :o