If a body of mass 12 kg is suspended by a string, find the horizontal force required to hold it at an angle of 30◦ from the vertical.?

Taking g = 10m/s² gives body’s weight = mg = 12*10 = 120N.

(You may need to use a different value for g, e.g. 9.8m/s².)

'B' is the body.

A

.

. .B.C

.

. .D

Arrow from B to D points down and is labelled ‘Weight, F (=120N)’.

Arrow from B to A points up at 30º to vertical and is labelled ‘Tension, T’.

Arrow from B to C points right and is labelled ‘Applied force, F’.

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Method 1.  Draw a vector addition triangle for the 3 forces.  This is a right-angled triangle:

P

.

Q R

The sides represent each force. PQ is weight, QR is F and RP is T.

Since angle QPR is 30º

F/W = QR/PQ = tan(30º)

F = Wtan(30º) = 120tan(30º) = 69N (rounded)

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Method 2.  Use components.

In the vertical direction, the vertical component of T balances the weight:

Tcos(30º) = 120

T = 120/cos(30º) = 138.6N

In the horizontal direction, the horizontal component of T balances F.

F = Tsin(30º) = 138.6sin(30º) = 69N (rounded)

F = m*g*sin 30°= 59 N