Question

One end of a meter stick is pinned to a table, so the stick can rotate freely in a plane parallel to the tabletop. Two forces, both parallel to the tabletop, are applied to the stick in such a way that the net torque is zero. The first force has a magnitude of 2.00 N and is applied perpendicular to the length of the stick at the free end. The second force has a magnitude of 6.00 N and acts at a 67.9° angle with respect to the length of the stick. Where along the stick is the 6.00-N force applied? Express this distance with respect to the end of the stick that is pinned. d= i

Answer 1

By torque balance about pinned point,

= 0 (given)

Tnet

7
1 =
7
2

here, 1 = torque due to first force(F1) = F1*r1*sinA1

7

2 = torque due to second force(F2) = F2*r2*sinA2

7

given, F1 = 2.00 N, F2 = 6.00 N

r1 = distance of F1 from pinned point = length of meterstick = 1 m

r2 = distance of F2 from pinned point = d = ??

A1 = angle between F1 and r1 = 90 deg (As, force acting perpendicular)

A2 = angle between F2 and r2 = 67.9 deg

then,

2.00*1*sin(90 deg) = 6.00*d*sin(67.9 deg)

d = 2/(6*sin(67.9 deg))

d = 0.360 m = 36.0 cm

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