Finding Reactions on Members Supported by Forces

[385sk117]

Homework Statement

Find the reactions exerted on the members by the supports.

Homework Equations

Sigma Fx = Net force acting on x-axis = 0
Sigma Fy = Net force acting on y-axis = 0
Sigma M = Net Moments = 0

The Attempt at a Solution

10N/m -> changes to 4N at centre acting downward so total force acting downward is 34N
the Sum of two y component of two forces from suppors then equals to 34N
If I set the regarding point to be the point on the left supports, then total moment is
30 x 0.2 + 4 x 0.6 - 50 - Fy x 0.8 where Fy is the y component of force from the right support. But this gives negative value for Fy which means that the supporting force from the right side support is downward. How can this be true?
Any problem with my steps?

I neglected Fx as there is no movement of bar and only force acting in x direction is the x component of supporting from left support and that should be equals to 0

 

[PhanthomJay]

Homework Statement

Find the reactions exerted on the members by the supports.

Homework Equations

Sigma Fx = Net force acting on x-axis = 0
Sigma Fy = Net force acting on y-axis = 0
Sigma M = Net Moments = 0

The Attempt at a Solution

10N/m -> changes to 4N at centre acting downward so total force acting downward is 34N
the Sum of two y component of two forces from suppors then equals to 34N
If I set the regarding point to be the point on the left supports, then total moment is
30 x 0.2 + 4 x 0.6 - 50 - Fy x 0.8 where Fy is the y component of force from the right support. But this gives negative value for Fy which means that the supporting force from the right side support is downward. How can this be true?
Any problem with my steps?

I neglected Fx as there is no movement of bar and only force acting in x direction is the x component of supporting from left support and that should be equals to 0

You are quite correct, a pin support can have vertical loads up or down, as long as the beam is pinned to it, and the support is pinned to the floor. The right support keeps it from rotating ccw. If the beam just rested on the right support, without a pin, or if the pin support just rested on the floor, it could not be in equilibrium. Sometimes the symbol for a roller support is misleading

 

[tiny-tim]

Hi 385sk117!

If I set the regarding point to be the point on the left supports, then total moment is
30 x 0.2 + 4 x 0.6 - 50 - Fy x 0.8 where Fy is the y component of force from the right support. But this gives negative value for Fy which means that the supporting force from the right side support is downward. How can this be true?

Your calculations look fine to me … the 50 N.m anticlockwise couple is easily more than the weights, so the net couple will be anticlockwise.

I notice that the "support" on the RHS is a different design to the one on the LHS … perhaps it's meant to be some sort of hook, that can exert a downward reaction?

 

[385sk117]

You are quite correct, a pin support can have vertical loads up or down, as long as the beam is pinned to it. The right support keeps it from rotating ccw. If the beam just rested on the right support, without a pin, it could not be in equilibrium.

Oh so the force is actually acting downwards on the right support? I thought that because it is the force that is supporting the bar, it must acting upwards.
Thanks

 

[PhanthomJay]

Oh so the force is actually acting downwards on the right support? I thought that because it is the force that is supporting the bar, it must acting upwards.
Thanks

Just watch your terminology: The right support exerts a downward force on the beam, and the beam exerts an upward force on the right support, per Newton 3. As Tiny Tim has noted, the support may be a hook of sorts, capable of exerting that downward force on the beam.