What direction is the force acting at A in this diagram?

[Benjamin_harsh]

Homework Statement

What direction is force at A in this diagram?

Relevant Equations

What direction is force at A in this diagram?

What direction is force acting at A in this diagram?

 

[haruspex]

Problem Statement: What direction is force at A in this diagram?
Relevant Equations: What direction is force at A in this diagram?

What direction is force acting at A in this diagram?

I don't understand your question.
Are you asking

• What force needs to be applied at A in order to achieve equilibrium?
• What force is being applied at A in the diagram? (answer, none)
• What net torque is being exerted at A by the forces in the diagram?
• Something else?

 

[Benjamin_harsh]

Just explain how this equation is written $\sum M_{A}$ = $F.(d + d') - F.d'$?

 

[jbriggs444]

Just explain how this equation is written $\sum M_{A}$ = $F.(d + d') - F.d'$?

As I read the equation, it is summing the "Moments" (also known as torques) about point A from the two applied forces.

The two moments are the left hand force $F$ multiplied by the moment arm whose perpendicular length is $d+d'$ and the right hand force $-F$ multiplied by the moment arm whose perpendicular length is $d'$.

At a guess, the text is trying to lead up to the notion of a "couple" -- a pair of forces that sum to zero but which exert the same non-zero net torque regardless of the reference point about which the torque is calculated.

 

[Benjamin_harsh]

The two moments are the left hand force $F$ multiplied by the moment arm whose perpendicular length is $d+d'$

Why did you multiply $F$ to both distances $d+d`$ instead of $d$ only?

 

[jbriggs444]

Why did you multiply $F$ to both distances $d+d`$ instead of $d$ only?

What is the perpendicular distance between point A and the line of action of the left hand force?

 

[Benjamin_harsh]

What is the perpendicular distance between point A and the line of action of the left hand force?

There is no physical contact between $A$ and $F$. So what perpendicular distance you are asking?

 

[jbriggs444]

There is no physical contact between $A$ and $F$. So what perpendicular distance you are asking?

The distance between the line of action of the left hand force and the reference point A.

We do this because it is the definition of torque.

Physical contact does not enter in. We are tasked with computing the sum of the torques from the two external forces. Both are computed relative to point A. Whether they are applied to a rigid body or not is irrelevant. The sum of the two will give a total torque. The torque will give a rate of change of angular momentum. The rate of change of angular momentum is still physically meaningful regardless of whether the torque is being applied to a rigid assembly.

 

[haruspex]

This thread has been superseded by https://www.physicsforums.com/threa...moments-for-this-diagram.972503/#post-6184580, where the question is explained better.

 

[Benjamin_harsh]

Now I understand. $\sum M_{A} = - F.(d + d') + F.d'$

Since it is acting anti clockwise, we need to multiply -ve to whole equation.

$\sum M_{A} = -(- F.(d + d') + F.d')$

$\sum M_{A} = F.d$

 

[haruspex]

Now I understand. $\sum M_{A} = - F.(d + d') + F.d'$

Since it is acting anti clockwise, we need to multiply -ve to whole equation.

$\sum M_{A} = -(- F.(d + d') + F.d')$

$\sum M_{A} = F.d$

Strictly speaking, the diagram is misleading. A force -F downwards is the same as a force F upwards. It should be shown either as F downwards or -F upwards.
E.g. if as F downwards then you would write the sum as F(d+d')-Fd and not need to flip the sign later.

Note that the answer does not depend on d'. This is an important result. A pair of forces that are equal and opposite as vectors but act along different lines form a "couple", and the torque of the couple is the same no matter what axis you choose.
This is the only situation in which the torque is independent of the choice of axis.

 

[Benjamin_harsh]

Strictly sp...hoice of axis.

But my method luckily worked. I am satisfied.

 

[haruspex]

But my method luckily worked. I am satisfied.

If you don't know how to it without trusting to luck you might not be so lucky next time.