What Angle Does Force F Make Relative to AD in Physics Problem Solving?

[cracktheegg]

Homework Statement

http://s1345.photobucket.com/user/Duk_Bato/media/Untitled2_zps52bc2429.png.html

Homework Equations

M= F * d

The Attempt at a Solution

I can't solve the last part the iii)

My problem is roller only got one force and how to resolve it?

It may sound excuses but, it will be best it someone can explain and even provide me with the working, since i can't understand really well

 

[cracktheegg]

Please clike the link, photobucket got error so i could not get the image link

 

[cracktheegg]

 

[SteamKing]

Homework Statement

http://s1345.photobucket.com/user/Duk_Bato/media/Untitled2_zps52bc2429.png.html

Homework Equations

M= F * d

The Attempt at a Solution

I can't solve the last part the iii)

My problem is roller only got one force and how to resolve it?

It may sound excuses but, it will be best it someone can explain and even provide me with the working, since i can't understand really well

We can't work out the problem for you; that is against the rules.

However, for the reaction force at D, assume that it has two components, Dx and Dy. You can still write equations of static equilibrium using Dx and Dy and solve for them.

 

[cracktheegg]

Dx= Fcos52
Dy= Fsin 52

?

 

[cracktheegg]

Can anyone briefly tell me what need to be done to solve this question, I have been trying but still wrong

 

[cracktheegg]

If I take A as the point where moment taken place and I separate D into force x and y and use the total moment= 0 can i solve?

 

[cracktheegg]

We can't work out the problem for you; that is against the rules.

However, for the reaction force at D, assume that it has two components, Dx and Dy. You can still write equations of static equilibrium using Dx and Dy and solve for them.

If I take A as the point where moment taken place and I separate D into force x and y and use the total moment= 0 can i solve?

 

[cracktheegg]

My moment equilibrium equation
Take A as moment:
5cos20(6)-5cos70(1)+2(cos20*8) - Fsin15(0.53) +F cos15 (0.55)=0

sin20*1.6=0.55
cos 2-*1.6=0.53

 

[haruspex]

My moment equilibrium equation
Take A as moment:
5cos20(6)-5cos70(1)+2(cos20*8) - Fsin15(0.53) +F cos15 (0.55)=0

A few problems there...
Check the signs. Draw the '5cos70' force. Which way does that act around A?
Resolving F first into horizontal and vertical and then into the 20 degree/70 degree system may be confusing. Perhaps simpler to skip the intermediate step and use 35/55 angles.
The parentheses are wrong in '2(cos20*8)'.

sin20*1.6=0.55
cos 2-*1.6=0.53

Haven't checked the numbers exactly, but I can see that can't be right. It makes cos 20 slightly less than sin 20; cos 20 must be significantly the larger.

 

[cracktheegg]

thanks, T T but I really can't understand the force of roller

what is the angle i should use?and the distance?

 

[haruspex]

thanks, T T but I really can't understand the force of roller

what is the angle i should use?and the distance?

If you resolve the normal force, F, at the roller into a component along AD and a component perpendicular to it then only one of those will have a moment about A.
What angle does AD make to the slope?
So what then is the angle F makes to AD?

 

[cracktheegg]

What angle does AD make to the slope? 35?
So what then is the angle F makes to AD? 20?

 

[cracktheegg]

isit Fsin35(12)?

 

[haruspex]

What angle does AD make to the slope? 35?

Yes.

So what then is the angle F makes to AD? 20?

F is at 90 degrees to the slope and AD is at 35 degrees to the slope. So what angle does F make to AD?

isit Fsin35(12)?

Close, in a sense. Try that again.