Physics

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1) The figure is below attached.

The torque of a force F with respect to a point O is by definition the vectorial product between the vectors representing the position vector of the force and the force itself. In its absolute value, the torque is

`|T| = |r xx F| = r*F*sin(alpha)`

To find angle `alpha` in the figure we write its tangent as:

`tan(alpha) =x/ y=2.3/1.4 rArr alpha =58.67 degree`

The magnitude of the position vector is

`r =sqrt(x^2+y^2) =sqrt(2.3^2+1.4^2) =2.69 m`

Therefore the magnitude of the force F is simply

`T = r*F*sin(alpha) rArr F = T/(r*sin(alpha)) =71/(2.69*sin(58.67)) =30.90 N `

Answer: the magnitude of the force applied is F=30.90 N

2)

From the above discussion results the the magnitude of the torque of the force applied is

`T =F*r*sin(beta) =16*0.63*sin(45) = 7.128 N*m`

Answer: the magnitude of the resulting torque is T =7.128 N*m

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