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- Homework Statement
- What is the net torque about an axis through point A?

- Relevant Equations
- Torque = F X d X sin theta

The net torque about an axis through point A is given by,

If I take the axis of rotation perpendicular to the paper and the solution I arrive would be the following below

Net torque = 30 cos45 x 1.5 - 10 cos30X 3

= 5.829Nm ( counterclockwise)

But the book gives an answer like this,

Net torque = 30 cos 45x1.5 - 10 x sin30 x 3 = 16.8Nm.

What did I miss? It looks the horizontal component is perpendicular to the axis of rotation.

If I take the axis of rotation perpendicular to the paper and the solution I arrive would be the following below

Net torque = 30 cos45 x 1.5 - 10 cos30X 3

= 5.829Nm ( counterclockwise)

But the book gives an answer like this,

Net torque = 30 cos 45x1.5 - 10 x sin30 x 3 = 16.8Nm.

What did I miss? It looks the horizontal component is perpendicular to the axis of rotation.