M1 question

A particle has mass 2kg. It is attached at B to the ends of two light inextensible strings AB and BC. When the particle hangs in equilibrium, AB makes an angle of 30 degrees with the vertical, as shown in diagram. The magnitude of the tension in BC is twice the magnitude of the tension in AB.

(a) Find, in degress to one decimal place, the size of the angle that BC makes with the vertical.

(b) Hence find, to 3 s.f., the magnitude of the tension in AB

[Edited to change a for angle to c to avoid confusion]

Unknown angle is c

Reolve horizontally

T(A) sin 30 = T (C) sin c and T (C) = 2 T(A) so

T(A) sin 30 = 2T (A) sin c

Cancel your T and you get sin c = 1/2 sin30 so c = 14.5, I think

Then resolving vertically should give you T

Reply 2

insparato

15

A particle has mass 2kg. It is attached at B to the ends of two light inextensible strings AB and BC. When the particle hangs in equilibrium, AB makes an angle of 30 degrees with the vertical, as shown in diagram. The magnitude of the tension in BC is twice the magnitude of the tension in AB.

(a) Find, in degress to one decimal place, the size of the angle that BC makes with the vertical.

(b) Hence find, to 3 s.f., the magnitude of the tension in AB

Tcos60 -2T sin(90-theta) = 0

Tcos60 = 2T(cos theta)

cos 60 = 2cos theta

cos60 = 2cos theta

1/4 = cos theta

theta = 75.5

So the angle to the vertical = 90-75.5 = 14.5

S

Resolving perpendicular

Tsin60 + 2Tsin75.5 - 2g = 0

T(sin 60 + 2sin75.5) = 2g

T = 19.6/sin 60 + 2sin75.5

T = 19.6/1.936 + 0.866

T = 19.6/2.802 = 7.0N

Reply 3

mike_J

OP

2

Thanks, yet again guys! :tongue:

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