The force of magnitude F acts along the edge of the triangular plate

[pyroknife]

Homework Statement

I attached the drawing.
The force of magnitude F acts along the edge of the triangular plate. Determine the moment of F about point O.

Homework Equations

M=rxF

The Attempt at a Solution

What I did was I just picked the top left point of the triangle and labeled that (0,h) which give me my r vector 0i+hj. After that I really don't know how to proceed except to find the angle by doing inverse tangent of h/b, but I just keep getting variable after variable. What other way is there to do this?

 

[stallionx]

Homework Statement

I attached the drawing.
The force of magnitude F acts along the edge of the triangular plate. Determine the moment of F about point O.

Homework Equations

M=rxF

The Attempt at a Solution

What I did was I just picked the top left point of the triangle and labeled that (0,h) which give me my r vector 0i+hj. After that I really don't know how to proceed except to find the angle by doing inverse tangent of h/b, but I just keep getting variable after variable. What other way is there to do this?

F is

b i - h j

what matters is the y direction for it is perpendicular and x is forming no moment

you just (bi) cross (-hj) to get ( -bh ) k, a vector of bh magnitude into the page.

 

[pyroknife]

The answer is supposed to be Fbh/(h^2+b^2)^.5. Answers do vary tho according to method, but I don't think yours matches this

 

[stallionx]

The answer is supposed to be Fbh/(h^2+b^2)^.5. Answers do vary tho according to method, but I don't think yours matches this

YES sorry I made a very Fundamental mistake and will try to rectify it


F= F *( b / ( h**2+b**2 ) ** 0.5 ) i + F*( h / ( h**2+b**2 ) ** 0.5 ) j

Fx * h = is the answer F*b*h *(h**2+b**2 ) ** -0.5

which is into the page

 

[pyroknife]

I'm still a little confused. I would like to do all moment problems by using M=rxF. How would that work in this problem. i can get the F vector but it's going to look really messy because I got to define the angle theta as tangent inverse of (b/h) then take the cosine of the tangent inverse of (b/h) to get the x component and same for the y.

 

[stallionx]

I'm still a little confused. I would like to do all moment problems by using M=rxF. How would that work in this problem. i can get the F vector but it's going to look really messy because I got to define the angle theta as tangent inverse of (b/h) then take the cosine of the tangent inverse of (b/h) to get the x component and same for the y.

M=r X F is already defined to be M= ( rx i + ry j + rz k ) X ( Fx i + Fy j + Fz k )

here simply r X ( Fx i + Fy j ) = r X Fx i + rXFy j by distribution property

but here : r= h j what do you get when you cross j by j ? for the latter part, of course a zero

here rXFyj drops and we are left with r X Fx i , here you cross jXi which points in the direction into the page.

 

[pyroknife]

M=r X F is already defined to be M= ( rx i + ry j + rz k ) X ( Fx i + Fy j + Fz k )

here simply r X ( Fx i + Fy j ) = r X Fx i + rXFy j by distribution property

but here : r= h j what do you get when you cross j by j ? for the latter part, of course a zero

here rXFyj drops and we are left with r X Fx i , here you cross jXi which points in the direction into the page.

its so hard to write all this online w/o subscripts and stuff. can r=b i too?

 

[stallionx]

its so hard to write all this online w/o subscripts and stuff. can r=b i too?

Sir,

r is starting from ( 0, 0 ) ---> take origin

and

draw a vector to the APPlication point of F.

r begins at ( 0,0 ) ends at application point of F.

 

[pyroknife]

Sir,

r is starting from ( 0, 0 ) ---> take origin

and

draw a vector to the APPlication point of F.

r begins at ( 0,0 ) ends at application point of F.

what I asked was basically the ending point can be any point on the line of action right?

 

[stallionx]

what I asked was basically the ending point can be any point on the line of action right?

No Sir, position vector starts at wherever you take the moment ( here (0,0) ) with respect to, and ends at starting point of F. ( which is h units above the (0,0)

 

[berkeman]

YES sorry I made a very Fundamental mistake and will try to rectify it


F= F *( b / ( h**2+b**2 ) ** 0.5 ) i + F*( h / ( h**2+b**2 ) ** 0.5 ) j

Fx * h = is the answer F*b*h *(h**2+b**2 ) ** -0.5

which is into the page

You need to stop doing the student's homework for them. You may provide hints, ask questions, find mistakes, etc. But it is against the PF rules to do the student's work for them.