Component of weight perpendicular to bar

[theone]

Homework Statement

http://postimg.org/image/tlvhadljz/
I am trying to find the component of weight perpendicular to the bar.

Homework Equations

The Attempt at a Solution

I tried as shown and I got mgcos(theta) but I am supposed to be getting mgsin(theta)

 

[Shreyas Shree]

Could you please explain the question. Even the picture seems vague.

 

[Delta2]

you are right in your calculations. mgsin(theta) is the component parallel to the bar. check this figure http://en.wikipedia.org/wiki/Inclined_plane#Analysis

 

[SammyS]

Homework Statement

http://postimg.org/image/tlvhadljz/
I am trying to find the component of weight perpendicular to the bar.

Homework Equations

The Attempt at a Solution

I tried as shown and I got mgcos(theta) but I am supposed to be getting mgsin(theta)

It's fairly basic trigonometry. The angle, θ, is obtuse, so you might be better off considering the angle π - θ which is acute. You can then more directly see that you need to use sin(π - θ) by considering opposite/hypotenuse.

 

[Shreyas Shree]

Homework Statement

http://postimg.org/image/tlvhadljz/
I am trying to find the component of weight perpendicular to the bar.

Homework Equations

The Attempt at a Solution

I tried as shown and I got mgcos(theta) but I am supposed to be getting mgsin(theta)

The theta you considered is not the same with the theta that is shown in the figure on the left.

 

[Delta2]

Oh dear they are right. theta is measured from the vertical to the ground plane (not from the ground plane itself) so the theta we considered is equal to the actual theta minus π/2. So the perpendicular component of weight is mgcos(theta-π/2)=mg(cos(theta)cos(pi/2)+sin(theta)sin(pi/2))=mgsin(theta).

 

[theone]

thanks everone