no need to calculate, B is doing the job unless you've never done such a thing before.

phr0n3:


It's a common sense question, but physics can provide answer too.
What of those that did not offer physics? So they will automatically fail the question?

Word!

Joe82834:
Well u can look at it from this angle. Both A and B are applying vertical upward forces so d one with d greater force should b d one bearing most of d grunt.

Assume distance from A to load is P n distance from B to load is Q(bearing in mind dat P is greater than Q). We can use d principle of moments about d given load.clockwise moment shd equal anticlockwise moment.
F_A × P. = F_B × Q
F_A = (Q/P) × F_B
Now since P is greater than Q it can b said that the force exerted at A is less than that at B

Joe's just thinking

I STRONGLY DISAGREE WITH YOU SIR AS THERE IS NO POSSIBLE AXIS OF ROTATION AT THE POINT OF APPLICATION OF THE LOAD.



To solve this problem, we must make two simplifications:
1.) The rod is uniform and massless(we lack information on its length)
2.) The rod rests on the shoulders of man A and B as though there is a knife edge placed there.

Obviously, the system(rod) is at equilibrium and we apply both conditions for equilibrium... Fnet =0 & Torque°net=0...
let the weight of the load be "W", man A and B provides reaction forces "Fa" and "Fb" respectively. let distance from manA to ManB be "x", distance from manA to load be "y" thus distance from load to manB is "(x-y)"...


since Fnet= 0,
Fa + Fb = W....(*)
since Torque -net = 0,
we consider torque about manB,
Fa*x = W×(x-y) ---(**)
we consider torque about manA,
Fb*x = W×y ---(***)
Note: Fb produces no Torque as its line of action passes through the axis of rotation in (**) likewise Fa in(***).

from(**),
Fa = [(x-y)/x]×W ... (I)
from(***),
Fb = [y/x]×W ....(II)

from figure in the problem, it is obvious that y>(x-y),
in view of (I) and (II), we can safely conclude that Fb > Fa. This means manB is carrying a heavier load.

Michael, dept. of Architecture, OAU ile-ife.

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TONYE001:
Mr. A is carrying the heavier load, I believe.

Why is this so?

The weight of the load and the weight of B are fairly at the same side.. (beyond the midpoint of the bar). The implication of this setup is that, the complex of B and the load would instigate a clockwise moment.

A direct consequence of this is that the bar at A would tend to be lifted up.. To prevent this from happening, Mr. A must send a force of the same magnitude as the weight of B and the load put together...this force must act in an opposite direction to the line of action of the weights of B and the load. This force would make up the anti-clockwise moment.

In conclusion, A is exerting a force equal to the weights of B and the load put together. Therefor, A is doing a greater work.

YOUR ANALYSIS IS TRUE IF THERE WAS SOME SORT OF SUPPORT AT THE MIDPOINT(which is not stated) THUS NO MOMENT ABOUT YOUR SO-CALLED MIDPOINT. MIND YOU, THE ROD CONTINUES TO THE RIGHT AFTER MANB.

XXX5:





I STRONGLY DISAGREE WITH YOU SIR AS THERE IS NO POSSIBLE AXIS OF ROTATION AT THE POINT OF APPLICATION OF THE LOAD.



To solve this problem, we must make two simplifications:
1.) The rod is uniform and massless(we lack information on its length)
2.) The rod rests on the shoulders of man A and B as though there is a knife edge placed there.

Obviously, the system(rod) is at equilibrium and we apply both conditions for equilibrium... Fnet =0 & Torque°net=0...
let the weight of the load be "W", man A and B provides reaction forces "Fa" and "Fb" respectively. let distance from manA to ManB be "x", distance from manA to load be "y" thus distance from load to manB is "(x-y)"...


since Fnet= 0,
Fa + Fb = W....(*)
since Torque -net = 0,
we consider torque about manB,
Fa*x = W×(x-y) ---(**)
we consider torque about manA,
Fb*x = W×y ---(***)
Note: Fb produces no Torque as its line of action passes through the axis of rotation in (**) likewise Fa in(***).

from(**),
Fa = [(x-y)/x]×W ... (I)
from(***),
Fb = [y/x]×W ....(II)

from figure in the problem, it is obvious that y>(x-y),
in view of (I) and (II), we can safely conclude that Fb > Fa. This means manB is carrying a heavier load.

Michael, dept. of Architecture, OAU ile-ife.

You made no point.

A

Before I make my submission or give a clue, I will like to ask my friend @XXX5 if he has heard of FBD?

First, well need to consider forces and the number of forces acting on the load.

There are four forces acting on the load and they are as follows:

*F Normal
*F Friction
*F Applied
*FGravity.

1. Applied Force
F app
An applied force is a force that is applied to an object by a person or another object. If a person is pushing a desk across the room, then there is an applied force acting upon the object. The applied force is the force exerted on the desk by the person.


2. Gravity Force
(also known as Weight)
F grav
The force of gravity is the force with which the earth, moon, or other massively large object attracts another object towards itself. By definition, this is the weight of the object. All objects upon earth experience a force of gravity that is directed "downward" towards the center of the earth. The force of gravity on earth is always equal to the weight of the object as found by the equation:
Fgrav = m * g
where g = 9.8 N/kg (on Earth)
and m = mass (in kg)
(Caution: do not confuse weight with mass.)


3. Normal Force
F norm
The normal force is the support force exerted upon an object that is in contact with another stable object. For example, if a book is resting upon a surface, then the surface is exerting an upward force upon the book in order to support the weight of the book. On occasions, a normal force is exerted horizontally between two objects that are in contact with each other. For instance, if a person leans against a wall, the wall pushes horizontally on the person.


4. Friction Force
F frict
The friction force is the force exerted by a surface as an object moves across it or makes an effort to move across it. There are at least two types of friction force - sliding and static friction. Though it is not always the case, the friction force often opposes the motion of an object. For example, if a book slides across the surface of a desk, then the desk exerts a friction force in the opposite direction of its motion. Friction results from the two surfaces being pressed together closely, causing intermolecular attractive forces between molecules of different surfaces. As such, friction depends upon the nature of the two surfaces and upon the degree to which they are pressed together. The maximum amount of friction force that a surface can exert upon an object can be calculated using the formula below:
Ffrict= µ • Fnorm

5. Air Resistance Force
F air
The air resistance is a special type of frictional force that acts upon objects as they travel through the air. The force of air resistance is often observed to oppose the motion of an object. This force will frequently be neglected due to its negligible magnitude (and due to the fact that it is mathematically difficult to predict its value). It is most noticeable for objects that travel at high speeds (e.g., a skydiver or a downhill skier) or for objects with large surface areas.


6. Tension Force
F tens
The tension force is the force that is transmitted through a string, rope, cable or wire when it is pulled tight by forces acting from opposite ends. The tension force is directed along the length of the wire and pulls equally on the objects on the opposite ends of the wire.

7. Spring Force
F spring
The spring force is the force exerted by a compressed or stretched spring upon any object that is attached to it. An object that compresses or stretches a spring is always acted upon by a force that restores the object to its rest or equilibrium position. For most springs (specifically, for those that are said to obey " Hooke's Law", the magnitude of the force is directly proportional to the amount of stretch or compression of the spring.


Solomon
Mr. Engineering 2012
H.N.D Mechanical Engineering Department
Federal Polytechnic Nasarawa.

O.N.D Mechanical Engineering Technology
National Metallurgical Training Institute Onitsha.


I'll be back soon.
React to these above!

laryom1:
lol. You should use moment of forces. Choose A and B as two fulcrums. Then calculate. U would see you are wrong

Yes ... that was what I actually meant..
thanks for the correction tho

.

dedons:
Which of these two is carrying the heavier load?

A or B?

since ur question come in lay man's form, the answer will therefore be provided with lay man's aproach noot scientifically because Op is surely not a sciencetist by profession. Hence the Load shifted towads B too much and A tends to add more salt to B injury when moving with excesive push consedering the Load is not at centre of gravity but liy on B Neck! So B is in pain thank you.

havocypie:



You made no point.

perhaps you may have to read the last paragraph

havocypie:
Before I make my submission or give a clue, I will like to ask my friend @XXX5 if he has heard of FBD?

First, well need to consider forces and the number of forces acting on the load.

There are four forces acting on the load and they are as follows:

*F Normal
*F Friction
*F Applied
*FGravity.

1. Applied Force
F app
An applied force is a force that is applied to an object by a person or another object. If a person is pushing a desk across the room, then there is an applied force acting upon the object. The applied force is the force exerted on the desk by the person.


2. Gravity Force
(also known as Weight)
F grav
The force of gravity is the force with which the earth, moon, or other massively large object attracts another object towards itself. By definition, this is the weight of the object. All objects upon earth experience a force of gravity that is directed "downward" towards the center of the earth. The force of gravity on earth is always equal to the weight of the object as found by the equation:
Fgrav = m * g
where g = 9.8 N/kg (on Earth)
and m = mass (in kg)
(Caution: do not confuse weight with mass.)


3. Normal Force
F norm
The normal force is the support force exerted upon an object that is in contact with another stable object. For example, if a book is resting upon a surface, then the surface is exerting an upward force upon the book in order to support the weight of the book. On occasions, a normal force is exerted horizontally between two objects that are in contact with each other. For instance, if a person leans against a wall, the wall pushes horizontally on the person.


4. Friction Force
F frict
The friction force is the force exerted by a surface as an object moves across it or makes an effort to move across it. There are at least two types of friction force - sliding and static friction. Though it is not always the case, the friction force often opposes the motion of an object. For example, if a book slides across the surface of a desk, then the desk exerts a friction force in the opposite direction of its motion. Friction results from the two surfaces being pressed together closely, causing intermolecular attractive forces between molecules of different surfaces. As such, friction depends upon the nature of the two surfaces and upon the degree to which they are pressed together. The maximum amount of friction force that a surface can exert upon an object can be calculated using the formula below:
Ffrict= µ • Fnorm

5. Air Resistance Force
F air
The air resistance is a special type of frictional force that acts upon objects as they travel through the air. The force of air resistance is often observed to oppose the motion of an object. This force will frequently be neglected due to its negligible magnitude (and due to the fact that it is mathematically difficult to predict its value). It is most noticeable for objects that travel at high speeds (e.g., a skydiver or a downhill skier) or for objects with large surface areas.


6. Tension Force
F tens
The tension force is the force that is transmitted through a string, rope, cable or wire when it is pulled tight by forces acting from opposite ends. The tension force is directed along the length of the wire and pulls equally on the objects on the opposite ends of the wire.

7. Spring Force
F spring
The spring force is the force exerted by a compressed or stretched spring upon any object that is attached to it. An object that compresses or stretches a spring is always acted upon by a force that restores the object to its rest or equilibrium position. For most springs (specifically, for those that are said to obey " Hooke's Law", the magnitude of the force is directly proportional to the amount of stretch or compression of the spring.


Solomon
Mr. Engineering 2012
H.N.D Mechanical Engineering Department
Federal Polytechnic Nasarawa.

O.N.D Mechanical Engineering Technology
National Metallurgical Training Institute Onitsha.


I'll be back soon.
React to these above!

I identified every one of those forces
we can ignore force of air resistance because the snapshot tells us the system is momentarily at rest...
we can ignore forces of static friction between the shoulders of both men because there is no tendency for motion of the rod.
the normal forces are my Fa and Fb. the applied force is W.
We cannot determine the force of gravity acting on the body... this is because there is no information on the length, mass and position of center of mass/gravity of the rod.

ERCROSS:
By the principles of physics
Load is measured in Kg, hence it can be measured with either spring balance or beams,though i think we can use one of the vector laws to calculate that, but the coordinates are not provided...
I think ur question should be "who is doing more work" , then we can use the formula for work done= Force * distance... In that case A is doing more work

U're without a doubt conclusive proof dat Education in Nigeria has failed, yes who is doing more work is right
but for the love of God isn't it common sense? It's B
Simple logic is if they're carrying an office desk, isn't the man carrying the side with the drawers that doing more than the man carrying the side without the drawer

ERCROSS:
By the principles of physics
Load is measured in Kg, hence it can be measured with either spring balance or beams,though i think we can use one of the vector laws to calculate that, but the coordinates are not provided...
I think ur question should be "who is doing more work" , then we can use the formula for work done= Force * distance... In that case A is doing more work

U're without a doubt conclusive proof dat Educational institutions in Nigeria has failed, yes who is doing more work is right
but for the love of God isn't it common sense? It's B
Simple logic is if they're carrying an office desk, isn't the man carrying the side with the drawers that doing more than the man carrying the side without the drawer

TONYE001:
Mr. A is carrying the heavier load, I believe.

Why is this so?

The weight of the load and the weight of B are fairly at the same side.. (beyond the midpoint of the bar). The implication of this setup is that, the complex of B and the load would instigate a clockwise moment.

A direct consequence of this is that the bar at A would tend to be lifted up.. To prevent this from happening, Mr. A must send a force of the same magnitude as the weight of B and the load put together...this force must act in an opposite direction to the line of action of the weights of B and the load. This force would make up the anti-clockwise moment.

In conclusion, A is exerting a force equal to the weights of B and the load put together. Therefor, A is doing a greater work.

Haaaaaaaaaaaaaa
U should really consider shooting all ur physics teachers or just abt anyone who has ever tot u anything in school.

It's common sense

actually, the snapshot only reveals an instance during which both men transported that rod together with the load. let us assume there was a net displacement of the rod from left to right. Both men have the same displacement vector (Da = Db = xî) in unit vector notation with respect to an origin at the lower left of the picture.
To talk about WORK-DONE by both men in the language of physics, we must have sufficient information on the horizontal component of the forces they applied, we lack such information. the forces of reaction and load DOES NO WORK as their line of action is perpendicular to the net displacement vector (xî).

In conclusion, based on the information given, we shouldn't talk about who is doing more work but who is applying a greater reaction force I.e. who feels the impact of the weight of the load the most. I have already proved man B is the man in question earlier on.

Some part of physics like this deals with common sense. Some people are here typing long issh and still coming out wrong.