# Reaction at pin support

## Homework Statement

I'm asked to find the W and I was told that the rope at C is in tension , there is reaction at B , my question is , is there any reaction at C ?

## The Attempt at a Solution

#### Attachments

• 107.3 KB Views: 289

Related Introductory Physics Homework Help News on Phys.org
haruspex
Homework Helper
Gold Member
2020 Award
Not sure I understand the question.
The rope AC applies a downward force at A through its tension. That is the only way that forces at C can affect forces at A. Ropes cannot transmit torque about their endpoints or forces perpendicular to the rope.

So, is there any reaction at c ?

SteamKing
Staff Emeritus
Homework Helper
So, is there any reaction at c ?
You don't care.

A free body diagram can be drawn around the beam which excludes C. All that matters, as far as the equilibrium of the beam is concerned, is that the rope is in tension, which means that Tc > 0.

You don't care.

A free body diagram can be drawn around the beam which excludes C. All that matters, as far as the equilibrium of the beam is concerned, is that the rope is in tension, which means that Tc > 0.
Why the reaction cant be drawn at c ?

SteamKing
Staff Emeritus
Homework Helper
Why the reaction cant be drawn at c ?
You don't care what the reaction at C is.

All you are interested in is finding the value of the distributed load w which keeps the rope in tension.

• goldfish9776
You don't care what the reaction at C is.

All you are interested in is finding the value of the distributed load w which keeps the rope in tension.
i have tried to do in this way , but i do not get the ans
vertical force = 80+10-RB+TC-RC-2W=0 --------equation 1
total moment about A = 80(1) +10(3)+W(2)(5) = 0
110+10W= 2RB , RB= (110+10W) / 2 ------------equation 2
total moment about B = -80(1)-2TC +2RC +10(1) +W(2)(3) =0

TC-RC = (6W-70) / 2 ----equation 3

Sub equation 2 and 3 into 1 ,
i gt 90-(110+10W) / 2 + (6W-70) / 2 -2W = 0
i gt 18W= 0
why cant i do int his way ?

SteamKing
Staff Emeritus
Homework Helper
i have tried to do in this way , but i do not get the ans
vertical force = 80+10-RB+TC-RC-2W=0 --------equation 1
total moment about A = 80(1) +10(3)+W(2)(5) = 0
110+10W= 2RB , RB= (110+10W) / 2 ------------equation 2
total moment about B = -80(1)-2TC +2RC +10(1) +W(2)(3) =0

TC-RC = (6W-70) / 2 ----equation 3

Sub equation 2 and 3 into 1 ,
i gt 90-(110+10W) / 2 + (6W-70) / 2 -2W = 0
i gt 18W= 0
why cant i do int his way ?
Looks messy.

Why don't you take moments about the pin at B? This will save you some work.

Remember, the reaction at C is not a load on the beam. The only load on the beam at point A is the tension in the rope, Tc.

Looks messy.

Why don't you take moments about the pin at B? This will save you some work.

Remember, the reaction at C is not a load on the beam. The only load on the beam at point A is the tension in the rope, Tc.
see it carefully , i did take the total moment about B ,
total moment about B = -80(1)-2TC +2RC +10(1) +W(2)(3) =0

if i ignore RC in my calculation , then my ans would be correct ?

SteamKing
Staff Emeritus
Homework Helper
see it carefully , i did take the total moment about B ,
total moment about B = -80(1)-2TC +2RC +10(1) +W(2)(3) =0

if i ignore RC in my calculation , then my ans would be correct ?
You haven't got any reasonable answer yet that I can see. Remember, the purpose of this exercise is to find the value of W which keeps the rope in tension.

Again, for the umteenth time, RC is not a load on the beam. Like haruspex said way back, you can't push on a rope. You haven't got any reasonable answer yet that I can see. Remember, the purpose of this exercise is to find the value of W which keeps the rope in tension.

Again, for the umteenth time, RC is not a load on the beam. Like haruspex said way back, you can't push on a rope. so , i have redo the question , here's what i gt :

80+10+2W -RB +TC= 0

moment about A = -80(1)+10(3) +W(2)(5) -2RB = 0
110+10W-2RB= 0
RB= (-110-10W) / 2

-80(1)+10(1) +2W(3) - TC(2) = 0
-70+6W-2TC= 0
2TC= -70+6W
TC = (-70 + 6W) / 2

90 + 2W - ((-110-10W) / 2 ) - ( (-70 + 6W) / 2 ) = 0
W=45N/m

is it correct ?

SteamKing
Staff Emeritus
Homework Helper
so , i have redo the question , here's what i gt :

80+10+2W -RB +TC= 0

moment about A = -80(1)+10(3) +W(2)(5) -2RB = 0
110+10W-2RB= 0
RB= (-110-10W) / 2
This is a superfluous calculation.

-80(1)+10(1) +2W(3) - TC(2) = 0
-70+6W-2TC= 0
2TC= -70+6W
TC = (-70 + 6W) / 2
The moment calculation about point B looks good.

90 + 2W - ((-110-10W) / 2 ) - ( (-70 + 6W) / 2 ) = 0
W=45N/m

is it correct ?
Then you went and spoiled it by adding the moments summed about point A.

You can write only one moment equation. Discard the moment equation about A.
Use the moment equation about B to find W, such that TC is always in tension. (TC > 0)

This is a superfluous calculation.

The moment calculation about point B looks good.

Then you went and spoiled it by adding the moments summed about point A.

You can write only one moment equation. Discard the moment equation about A.
Use the moment equation about B to find W, such that TC is always in tension. (TC > 0)
so , the W = 60/ 7= 8.57???

SteamKing
Staff Emeritus
Homework Helper
so , the W = 60/ 7= 8.57???
Where did this come from?

Where did this come from?
from the moment about B above

SteamKing
Staff Emeritus