Calculate the barbell acceleration while being lifted

[Jeff97]

Homework Statement

This diagram shows a weightlifter in the initial
stages of his lift. He lifts a weighted bar which
has a 70 kg mass disc on each end and the cross
bar has a mass of 10 kg.
Given g = 9.8 N kg!!
b. Calculate the total weight that he is lifting.

c. During the lift, one of his hand applies a force of 795 N and the other applies a force of
810 N. Calculate the size of the acceleration of the weights during the lift.


The weightlifter asks his assistant to add
more weight to the bar. The assistant lifts
the left end first and adds more weight.
He then lifts the right end and does the
same – as shown in the diagram.
This sketch is a simplified diagram with the forces and distances.

d. (i) Use the information in the diagram above to calculate the size of the force used to lift the
right end of the bar.

Relevant Equations

Don't know.

 

[berkeman]

You need to show the Relevant Equations and your work on the solutions before we can offer tutorial help. Also:

Given g = 9.8 N kg!

Are you sure about those units? Remember that F=ma...

 

[Jeff97]

for b isn't he just lifting 70kg+70kg+10kg=150kg?

for c I work out each hand separately? giving me 795N/150kg=5.3m/s^2 for the righthand and 810N/150kg=5.4m/s^2

For part b I could be wrong because I haven't considered gravity?

For part c I could be wrong because of part b or because Each hand may not be lifting 150kg they may be lifting 75kg? half of the weight each?

 

[doggydan42]

for b isn't he just lifting 70kg+70kg+10kg=150kg?

It's asking for weight. Think about the relation between mass and weight. Also, consider units to determine what you just calculated.

 

[Jeff97]

do you mean 150x9.8=1470N

 

[doggydan42]

do you mean 150x9.8=1470N

I do mean that, but why? You should make it clear what quantity you are calculating and how it relates to weight.

 

[doggydan42]

but is it not the answer to b?

Why did you multiply 150 by 9.8, where did you get those numbers and how does it relate to the weight?

Also, it's best to keep track of units, being lazy and dropping units mid-calculation can make it hard to figure out units later; it can also cost you some points depending on your teacher.

 

[doggydan42]

150kg was multiplied by 9.8 N Kg^-1 to convert 150kg to Newtons

I would not call it a conversion. You are calculating Newtons which represents a force. The force depends on the acceleration, so different values of a in F = ma will give you different forces for the same mass. I recommend writing it as follows:
$$F = ma = (\# kg)(\# \frac{m}{s^2}) = \# N$$.

Why are you using 9.8 m/s^2 for the force and how does the quantity you are calculating (the force) relate to the weight? I can see you are using a = g, so it might be more appropriate to say $F = mg$ but why g for the weight?

 

[Jeff97]

You're now confusing me F=mg where F stands for Force, m stands for mass, g stands for gravity (9.8)

So In theroy, you'd do F=mg (mass being 70kg+70kg+10kg) 150kg/(divied by) 9.8(g)=1470N(Newtons

 

[doggydan42]

You're now confusing me F=mg where F stands for Force, m stands for mass, g stands for gravity (9.8)

So In theroy, you'd do F=mg (mass being 70kg+70kg+10kg) 150kg/(divied by) 9.8(g)=1470N(Newtons

Sorry if anything was confusing, I can clarify something if you let me know. That looks good. You calculated the force, it's the same thing you did earlier, just in that line you specified more so where you were getting that calculation. Now what does that force have to do with weight? Once you make the connection you should find your answer to b.

 

[Orodruin]

For c, what is the total force acting on the bar? How does that total force relate to the acceleration of the bar?

a weighted bar which
has a 70 kg mass disc on each end and the cross
bar has a mass of 10 kg.

This makes me wonder if the person writing the question has ever set foot in a gym ...

 

[doggydan42]

The force acting on the bar? would be gravity? it relates to accelerating because it makes it harder to accelerate? For calculating the acceleration do I do each hand individually?

I think Orodruin is referring to calculations and not conceptual. So when he/she is asking what is the total force, he/she is asking you to calculate the total force. And then what is the mathematical relation of force and acceleration.

For the calculations, you are looking for the total force. So take into account all the forces acting on the bar.

 

[Jeff97]

Now, what does that force have to do with weight? Once you make the connection you should find your answer to b.

So for b is the total weight lifted not 150kg? For c, do I use a=f/m force being 810N+795N=1605N / 150kg a=10.7m/s^-1 ( I understand this could be wrong, I'd like to know how I'm going wrong?

 

[doggydan42]

So for b is the total weight lifted not 150kg? For c, do I use a=f/m force being 810N+795N=1605N / 150kg a=10.7m/s^-1 ( I understand this could be wrong, I'd like to know how I'm going wrong?

For b, 150 kg is a mass. Mass is not the same as weight, so you still need to solve for the weight using that mass.

For c, you need the net force. You calculated the force the lifter is using to lift the bar, but there is still another force you are not accounting for.

 

[Orodruin]

a=10.7m/s^-1

This is wrong not only because you did not use the total force, as mentioned in the previous post, but you also have gotten the units wrong. m/s is a unit of velocity, m/s^-1 = m s is a unit of something entirely different, and m/s^2 is a unit of acceleration.

 

[Jeff97]

So for B he lifts a total weight of 1470N? and for C, in the formula a=f/m I believe I use 150kg. But am Uncertain as what to use for the force F. Do I use 795N...810N (these added up 1605?) or 1470N ( or 1605+1470N?)

 

[Orodruin]

Do I use 795N...810N (these added up 1605?) or 1470N ( or 1605+1470N?)

Direction of forces is important. You want the total force, i.e., the sum of all forces on the barbell with directions taken into account.

 

[Jeff97]

All forces are taken into account...I suppose the weight of the bar is 1470N (acting downwards) and he is pulling upwards with a force of 1605N combined? So I can rule out - because you said sum meaning addition. So on this, I think it's 1605N+1470N=3075N?(Note: we are only dealing with the first diagram the second diagram is for question d.)

 

[jbriggs444]

So I can rule out - because you said sum meaning addition

@Orodruin also said:

with directions taken into account.

If one simply adds magnitudes, is that taking directions into account?

 

[Orodruin]

All forces are taken into account...I suppose the weight of the bar is 1470N (acting downwards) and he is pulling upwards with a force of 1605N combined? So I can rule out - because you said sum meaning addition. So on this, I think it's 1605N+1470N=3075N?(Note: we are only dealing with the first diagram the second diagram is for question d.)

By this reasoning, an object sitting on a bookshelf would accelerate at a rate of 19.6 m/s^2.

 

[Jeff97]

Ok, I gather you take 1605-1470=135N This seems to take into account of direction. a=f/m a=135/150=0.9m/s^2

 

[doggydan42]

Ok, I gather you take 1605-1470=135N This seems to take into account of direction. a=f/m a=135/150=0.9m/s^2

That looks right.

For part d, you should use torque to calculate the lifting force.

 

[Jeff97]

I have 850N,100N and 750N There distance from the pivots are 850Nx0.7m =595Nm. 100Nx1.4m.=140Nm 750Nx2.4 =1800Nm (Could be wrong if I misread the distances on the diagram) what's the next step?

 

[doggydan42]

I have 850N,100N and 750N There distance from the pivots are 850Nx0.7m =595Nm. 100Nx1.4m.=140Nm 750Nx2.4 =1800Nm (Could be wrong if I misread the distances on the diagram) what's the next step?

You need to calculate the net torque. Account for direction. Also, verify you used the correct r.

 

[Jeff97]

clockwise movemnets, =(850x0.7)+(100x1.4)
=595+140
=735Nm

Anticlockwise movements = (750x2.4)= 1800Nm

Ffarright(x)2.8

clockwise movements=anti clockwise movements. plank is at equalibrium

Im stuck

 

[doggydan42]

clockwise movemnets, =(850x0.7)+(100x1.4)
=595+140
=735Nm

Anticlockwise movements = (750x2.4)= 1800Nm

Look at the diagram carefully. You are not using the correct distances (r-values). Also pay attention to the direction of the force.

 

[Jeff97]

Is the 850N no distance away from the pivot? The 100N appears to be 0.7+0.7 meters away, the 750N also appears to be 1.2+1.2 meters away from the pivot. All the forces I have stated are acting downwards.

 

[doggydan42]

Is the 850N no distance away from the pivot? The 100N appears to be 0.7+0.7 meters away, the 750N also appears to be 1.2+1.2 meters away from the pivot. All the forces I have stated are acting downwards.

The distances for each force is given in the diagram, you shouldn't have to calculate them. There is still one more force, the one you are calculating.

 

[doggydan42]

So (850x0)+(750x0.7)= 920Nm for clockwise.

Anti-clockwise (750x1.2)=93Nm (do i take 920-93?) or just do lifting force (Fx1.4)

From the diagram, it looks like there are 3 forces in the same direction, and one in the other direction. Where are you getting 93 Nm in the anti-clockwise direction?

 

[Jeff97]

Clockwise=(850x0)+(100x0.7)+(750x1.2)=850+70+900=1820Nm

anticlockwise= fx1.4= 1820/1.4=1300N?

 

[doggydan42]

Clockwise=(850x0)+(100x0.7)+(750x1.2)=850+70+900=1820Nm

anticlockwise= fx1.4= 1820/1.4=1300N?

$f*1.4 \neq 1820/1.4$

Make each step clear, and keep track of units.

 

[doggydan42]

True, f=3250/3 f=1083N

Where are you getting 3250? Your had the right idea last time. Just make it clear and write down units at each step.

 

[doggydan42]

The answer is 1300N, am I not right?

That's not right. Show your work clearly.

 

[Jeff97]

(850x0)+(100x0.7)+(750x1.2)=970Nm

Fb x 1.4 970/1.4= 692NIS this correct?

 

[doggydan42]

Fb x 1.4 970/1.4= 692N

Check your rounding.