# how to work out the reaction force on a uniform rod??

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so i know that the sum of the anticlockwise = sum of the clockwise moments, but how to work out the reaction force for a uniform beam with two reaction forces?

in this case we have two unknowns and im not sure how to go about it :/

in this case we have two unknowns and im not sure how to go about it :/

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(Original post by

so i know that the sum of the anticlockwise = sum of the clockwise moments, but how to work out the reaction force for a uniform beam with two reaction forces?

in this case we have two unknowns and im not sure how to go about it :/

**Mimi9335**)so i know that the sum of the anticlockwise = sum of the clockwise moments, but how to work out the reaction force for a uniform beam with two reaction forces?

in this case we have two unknowns and im not sure how to go about it :/

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**Mimi9335**)

so i know that the sum of the anticlockwise = sum of the clockwise moments, but how to work out the reaction force for a uniform beam with two reaction forces?

in this case we have two unknowns and im not sure how to go about it :/

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(Original post by

Please post the whole question.

**Notnek**)Please post the whole question.

so the question is this:

AB is a uniform rod of length 5m and weight 20N,. In these diagrams, AB is resting in a horizontal position on supports C and D. Find the magnitude of the reaction at C and D.

distance from A to C (C is a support): 1.5m

distance from C to D (another support): 2.7m

distance from D (the support), to B (the end of the rod): 0.8m

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(Original post by

yup, sorry about that

so the question is this:

AB is a uniform rod of length 5m and weight 20N,. In these diagrams, AB is resting in a horizontal position on supports C and D. Find the magnitude of the reaction at C and D.

distance from A to C (C is a support): 1.5m

distance from C to D (another support): 2.7m

distance from D (the support), to B (the end of the rod): 0.8m

**Mimi9335**)yup, sorry about that

so the question is this:

AB is a uniform rod of length 5m and weight 20N,. In these diagrams, AB is resting in a horizontal position on supports C and D. Find the magnitude of the reaction at C and D.

distance from A to C (C is a support): 1.5m

distance from C to D (another support): 2.7m

distance from D (the support), to B (the end of the rod): 0.8m

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(Original post by

Got it?

**TastyChicken**)Got it?

lol sorry for this hassle but its really bugging me

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#7

(Original post by

im still trying to figure out how to apply the idea you mentioned, how would i get to the simultaneous equations bit without finding the moments first?

lol sorry for this hassle but its really bugging me

**Mimi9335**)im still trying to figure out how to apply the idea you mentioned, how would i get to the simultaneous equations bit without finding the moments first?

lol sorry for this hassle but its really bugging me

Use the two rules to come up with two equations that contain the reaction force for C and D. Then you can apply simultaneous equations to solve it.

First set A as a pivot and calculate the equilibrium of moments, namely clockwise moments must equal anti-clockwise moments. Then use of the rule that all upward forces namely C and D must equal all downward forces ie. 20.

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#8

**Mimi9335**)

yup, sorry about that

so the question is this:

AB is a uniform rod of length 5m and weight 20N,. In these diagrams, AB is resting in a horizontal position on supports C and D. Find the magnitude of the reaction at C and D.

distance from A to C (C is a support): 1.5m

distance from C to D (another support): 2.7m

distance from D (the support), to B (the end of the rod): 0.8m

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(Original post by

There are two reaction forces, not just one, for the two points of contact. If we label these reaction forces as R (at C) and S at (at D) then we have moments to consider. If you take moments about point C, then you will end up with one equation and a single variable because R can be completely ignored. Hence you find S. Secondly, you can consider the fact that the reaction forces must sum to the weight force which means that R+S=20

**RDKGames**)There are two reaction forces, not just one, for the two points of contact. If we label these reaction forces as R (at C) and S at (at D) then we have moments to consider. If you take moments about point C, then you will end up with one equation and a single variable because R can be completely ignored. Hence you find S. Secondly, you can consider the fact that the reaction forces must sum to the weight force which means that R+S=20

(Original post by

No worries. Have you drawn out the problem, this often helps visualize what is happening?

Use the two rules to come up with two equations that contain the reaction force for C and D. Then you can apply simultaneous equations to solve it.

First set A as a pivot and calculate the equilibrium of moments, namely clockwise moments must equal anti-clockwise moments. Then use of the rule that all upward forces namely C and D must equal all downward forces ie. 20.

**TastyChicken**)No worries. Have you drawn out the problem, this often helps visualize what is happening?

Use the two rules to come up with two equations that contain the reaction force for C and D. Then you can apply simultaneous equations to solve it.

First set A as a pivot and calculate the equilibrium of moments, namely clockwise moments must equal anti-clockwise moments. Then use of the rule that all upward forces namely C and D must equal all downward forces ie. 20.

thank you so much for your help, guys honestly I really appreciate it! May you have all the full marks in the world

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