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 = [tex]\frac{360+240}{450}[/tex]
tan θ = [tex]\frac{600}{450}[/tex]
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:
[tex]M__B}=(Fcos \theta *240)-(Fsin \theta *450)[/tex]
where Force(F) = Force in the cord AC = 1350 N and θ = 53.13° ; we have:
[tex]M__B}=(1350&cos 53.13^0 *240)-(1350sin 53.13^0 *450)[/tex]
[tex]M__B}= 194400.463-485999.348[/tex]
[tex]M__B}=-291598.885 N-mm\\[/tex]
[tex]M__B}=-291.598 N-m[/tex]
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 [tex](M__B})[/tex],
we have:
[tex]M__B}=-Fcos \theta *360 - Fsin \theta * 0[/tex]
[tex]M__B}=-Fcos \theta *360 - 0[/tex]
[tex]M__B}=-Fcos \theta *360[/tex]
where Force(F) = 1350 N and θ = 53.13° ; we have:
[tex]M__B}= -1350*cos53.13^0*360[/tex]
[tex]M__B}= -291600 N-mm[/tex]
[tex]M__B}= -291.6 N-m[/tex]
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
The moment of a force about a point is given by the product of the force and the perpendicular distance of the point from the line of action of the force
(a) The moment about B, with force acting at A, is approximately 291.598 N-m
(b) The moment about point B with force acting at C is approximately 291.60069 N-m
The reason the above values are correct are as follows:
The given parameters are;
The cord AC holds the rod AB in place
The tension in the cord AC = 1,350 N
The length of segment, c = 360 mm
(a) The angle the cord makes with the horizontal, θ is give as follows;
[tex]\theta = arctan\left( \dfrac{240 + 360}{450} \right) \approx 53.13^{\circ}[/tex]
The moment about point B due to the tension in the cord, is given as follows;
[tex]M_B = 1,350 \times sin(53.13^\circ}) \times 450 - 1,350\times cos (53.13^\circ}) \times 240 \approx 291598.9[/tex]
Therefore, the moment about B, [tex]M_B[/tex] ≈ 291.598 N-m
(b) To find the moment about B, though C
Taking moment about B gives
[tex]M_B = 1,350 \times cos(53.13^\circ}) \times 360+ 1,350\times sin(53.13^\circ}) \times 0 \approx 291,600.69[/tex]
The moment about point B through point C, [tex]M_B[/tex] ≈ 291.60069 N-m
Learn more about finding the moment of a force here:
https://brainly.com/question/24606440
