Question

Rod AB is held in place by the cord AC. Knowing that the tension in the cord is 1350 N and that c 5 360 mm, determine the moment about B of the force exerted by the cord at point A by resolving that force into horizontal and vertical components applied (a) at point A, (b) at point C.

Answer 1

Answer:

291.598 N-m

291.6 N-m

Explanation:

Let's first take a  look at the free bodily diagrammatic representation.

The first diagram will aid us in answering  question (a), so as the second diagram will facilitate effective understanding when solving for question (b).

Let's first determine our angle θ from the diagram

To find angle θ ; we have :

tan θ  = $\frac{360+240}{450}$

tan θ  = $\frac{600}{450}$

tan θ  = 1.333

θ  = tan⁻¹ (1.333)

θ  = 53.13°

Now, to determine the moment about B of the force exerted by the cord at point A by resolving that force into horizontal and vertical components applied at point A.

We have:

$M__B}=(Fcos \theta *240)-(Fsin \theta *450)$

where Force(F) = Force in the cord AC = 1350 N and θ  = 53.13° ; we have:

$M__B}=(1350&cos 53.13^0 *240)-(1350sin 53.13^0 *450)$

$M__B}= 194400.463-485999.348$

$M__B}=-291598.885 N-mm$

$M__B}=-291.598 N-m$

Since the negative sign illustrates just the clockwise movement ; then the moment about B of the force exerted by the cord at point A by resolving that force into horizontal and vertical components applied at point A = 291.598 N-m

b) From the second diagram, taking the moment at point B $(M__B})$,

we have:

$M__B}=-Fcos \theta *360 - Fsin \theta * 0$

$M__B}=-Fcos \theta *360 - 0$

$M__B}=-Fcos \theta *360$

where Force(F) =  1350 N and θ  = 53.13° ; we have:

$M__B}= -1350*cos53.13^0*360$

$M__B}= -291600 N-mm$

$M__B}= -291.6 N-m$

Since the negative sign illustrates just the clockwise movement ; then the moment about B of the force exerted by the cord at point A by resolving that force into horizontal and vertical components applied at point C = 291.6 N-m