Respuesta :
Answer:
The moment is 81.102 k N-m in clockwise.
Explanation:
Given that,
Force = 260 N
Side = 580 mm
Distance h = 370 mm
According to figure,
Position of each point
[tex]O=(0,0)[/tex]
[tex]A=(0,-b)[/tex]
[tex]B=(h,0)[/tex]
We need to calculate the position vector of AB
[tex]\bar{AB}=(h-0)i+(0-(-b))j[/tex]
[tex]\bar{AB}=hi+bj[/tex]
We need to calculate the unit vector along AB
[tex]u_{AB}=\dfrac{\bar{AB}}{|\bar{AB}|}[/tex]
[tex]u_{AB}=\dfrac{h\hat{i}+b\hat{j}}{\sqrt{h^2+b^2}}[/tex]
We need to calculate the force acting along the edge
[tex]\hat{F}=F(u_{AB})[/tex]
[tex]\hat{F}=F(\dfrac{h\hat{i}+b\hat{j}}{\sqrt{h^2+b^2}})[/tex]
We need to calculate the net moment
[tex]\hat{M}=\hat{F}\times OA[/tex]
Put the value into the formula
[tex]\hat{M}=F(\dfrac{h\hat{i}+b\hat{j}}{\sqrt{h^2+b^2}})\times(-b\hat{j})[/tex]
[tex]\hat{M}=\dfrac{F}{\sqrt{h^2+b^2}}((h\hat{i}+b\hat{j})\times(-b\hat{j}))[/tex]
[tex]\hat{M}=\dfrac{F}{\sqrt{h^2+b^2}}(-bh\hat{k})[/tex]
[tex]\hat{M}=-\dfrac{bhF}{\sqrt{h^2+b^2}}[/tex]
Put the value into the formula
[tex]\hat{M}=-\dfrac{580\times10^{-3}\times370\times10^{-3}\times260}{\sqrt{(370\times10^{-3})^2+(580\times10^{-3})^2}}[/tex]
[tex]\hat{M}=-81.102\ \hat{k}\ N-m[/tex]
Negative sign shows the moment is in clockwise.
Hence, The moment is 81.102 k N-m in clockwise.
The moment will be 81.102 k N-m in a clockwise direction. The moment is used to rotate or twist the object.
What is a moment?
The moment is defined as the product of the force and the perpendicular distance from the pivot point. Its unit is KN-m.
The given data in the problem is;
F is the Force = 260 N
b is the Side = 580 mm
h is the distance = 370 mm
Position of the points is found by;
O(0,0)
A(0,-b)
B(h,0)
The position vector for the AB will be;
[tex]\vec AB = (h-0)+ (0-(-b))j \\\\ \vec AB =h \vec i + b \vec j[/tex]
The unit vector along with AB
[tex]\rm u_AB = \frac{\vec AB }{|\vec AB|} \\\\ \rm u_AB = \frac{h \vec i + b \vec j}{\sqrt{h^2+b^2} }[/tex]
The net moment is found by;
[tex]\hat M = \hat F \times OA \\\\ \hat M =F \frac{h \vec i + b \vec j}{\sqrt{h^2+b^2} }\times (b \vec j) \\\\ \hat M =\frac{F}{\sqrt{h^2+b^2}} \times (bh \hat k) \\\\ \hat M =- \ \frac{bhf}{\sqrt{h^2+b^2}}[/tex]
[tex]\hat M =- \ \frac{bhf}{\sqrt{h^2+b^2}} \\\\ \hat M =- \ \frac{580 \times 10^-3 \times 370 \times 10^-3 \times 260 }{\sqrt{(370\times 10^-3)^2+(580\times 10^-3)^2}} \\\\ \hat M =- 81.102 \ KNm[/tex]
-ve sign shows that moment is clockwise.
Hence the moment will be 81.102 k N-m in a clockwise direction.
To learn more about the moment refer to the link;
https://brainly.com/question/6278006
