Painter on a crate - Acceleration

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Homework Statement
Please Refer the Image below
Relevant Equations
F=ma
Here's my attempt to solve part b. I am not sure how to approach part a.
Please be kind to help me.
The answer of part be must be 2m/s² but I get a different answer.
 
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  • Printed textbook question 31 with two parts: (a) explain why the load on the back wheels of a car increases when accelerating, (b) a painter in a crate hangs beside a building; he weighs 100 kgf, pulls the rope and exerts 45 kgf on the crate floor; crate weighs 25 kgf; find the acceleration.
    SmartSelect_20220416-091325_Drive.jpg
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  • Handwritten force diagram of a pulley system with a 45 kgf block and a 100 kgf load, tension T, and handwritten equations (1000 + 450 - T = 100a, 450 - T = 25a) solved for a = 100/75.
    20220416_091351.jpg
    32.8 KB · Views: 128
  • Line drawing of a painter standing on a crate-shaped platform, holding a rope that runs up over a pulley at the top and back down to the platform.
    SmartSelect_20220416-092312_Drive.jpg
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Last edited:
Have you tried a FBD and summation of forces and moments for part a?
 
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Lnewqban said:
Have you tried a FBD and summation of forces and moments for part a?
I am not sure how to start on that. Could you please give me an example?
 
Shreya said:
I am not sure how to start on that. Could you please give me an example?
Draw a side view of the car showing a front wheel and a rear wheel.
Show the vertical and horizontal forces the ground exerts on the wheels, assigning variable names to them, bearing mind that the car is accelerating.
What force and torque balance equations can you write?
 
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Shreya said:
I am not sure how to start on that. Could you please give me an example?
Your FBD is not right. You are forgetting forces and the signs don't look right. And there are no units. If you plug the numbers in you must also use the correct units.

I suggest you take up to be positive and write down the force equations for the man and the lift. Try to get those correct first.

Use Latex if you can:
$$Ma = \dots$$$$ma = \dots$$
 
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PeroK said:
Your FBD is not right. You are forgetting forces and the signs don't look right. And there are no units. If you plug the numbers in you must also use the correct units.

I suggest you take up to be positive and write down the force equations for the man and the lift. Try to get those correct first.

Use Latex if you can:
$$Ma = \dots$$$$ma = \dots$$
Post #3 was in respect of part a, the question about an accelerating car.
The attempt attached in post #1 was for part b. The OP seems to have taken the acceleration as positive down. All your points are valid but the answer obtained looks ok.
 
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haruspex said:
Post #3 was in respect of part a, the question about an accelerating car.
The attempt attached in post #1 was for part b. The OP seems to have taken the acceleration as positive down. All your points are valid but the answer obtained looks ok.
The force equation for the lift omits the weight of the lift itself. The answer should be ##0.2g##, which is what I get and is the answer quoted by the OP.
 
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PeroK said:
The force equation for the lift omits the weight of the lift itself. The answer should be ##0.2g##, which is what I get and is the answer quoted by the OP.
Ahem... yes, I missed that.
 
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PeroK said:
force equation for the lift omits the weight of the lift itself
I am so sorry, I have included it in the new solution.
Thanks a lot for your help! ☺

For the Crate:
$$T - 450N - 250N = 25a$$
For the Painter:
$$T+450N - 1000N = 100a$$
On solving I get, ##a =2m/s²##
##T = 75kgf##
I tried latex for the first time! It works!
 
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haruspex said:
Draw a side view of the car showing a front wheel and a rear wheel
I tried it. Here's what I found. Correct me if I am wrong.
The force due to the ground on the wheels cause the acceleration.
They also cause a torque on the car making it rise up, like '/'
Initially the reaction forces due to the 4 wheels balanced the force of gravity. Now the back wheels have to bear the entire load.
The torque due to the reaction force - brings the car back down
 
Shreya said:
I tried it. Here's what I found. Correct me if I am wrong.
The force due to the ground on the wheels cause the acceleration.
They also cause a torque on the car making it rise up, like '/'
Initially the reaction forces due to the 4 wheels balanced the force of gravity. Now the back wheels have to bear the entire load.
The torque due to the reaction force - brings the car back down
Yes, the difference in the normal forces creates a torque, balancing the torque from the frictional forces.
 
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