For the lovers of Physics

[Mokaze]

 

[Mokaze]

Is Physics the hardest subject of all??!! -----Unbelievable !!! 🤣

 

[Teknocrat]

mg−T=ma (object to the right of the pulley)

Tension is positive, mg is negative as follows
T−mg=ma

4T(pulleys)-2mg(downward force)=2ma(upward force)

Also T=mg/2(force equal half of its weight)

Wenye kujua maths, wata substitute T kwenye formula 4T-2mg=2ma
a=?
Inaonekana a man of mass (m) hana ubavu huo.......

 

[Mokaze]

mg−T=ma (object to the right of the pulley)

Tension is positive, mg is negative as follows
T−mg=ma

4T(pulleys)-2mg(downward force)=2ma(upward force)

Also T=mg/2(force equal half of its weight)

Wenye kujua maths, wata substitute T kwenye formula 4T-2mg=2ma
a=?
Inaonekana a man of mass (m) hana ubavu huo.......

Ubavu wa kufanya nini???

The question is what will be his upward acceleration ??

Above all you have tried a great deal, a big up ✔✔✔.

 

[Abrianna]

Everything is Energy. Einstein’s 1905 formula E = mc2 explains the relationship between Energy and matter, i.e., that Energy and matter are interchangeable – that, in reality, everything is Energy – dancing, fluid, ever-changing Energy.

 

[simplemind]

What if we consider only boundary forces- 4 T upwards where t is tension force in rope and 2mg downward gravitational force.
By Newtons second law of motion F=ma: 4T-2mg=2ma . 2,unknowns(T &a) ,one equation CHECKMATE

 

[Abrianna]

F=GMm /r2

where F-represents the force in Newtons, M and m represent the two masses in kilograms, and r represents the separation in meters. G represents the gravitational constant, which has a value of 6.674⋅10−11N(m/kg)2. Because of the magnitude of G, gravitational force is very small unless large masses are involved.

 

[Teknocrat]

Ubavu wa kufanya nini???

The question is what will be his upward acceleration ??

Above all you have tried a great deal, a big up ✔✔✔.

a=0

 

[Andres]

mg−T=ma (object to the right of the pulley)

Tension is positive, mg is negative as follows
T−mg=ma

4T(pulleys)-2mg(downward force)=2ma(upward force)

Also T=mg/2(force equal half of its weight)

Wenye kujua maths, wata substitute T kwenye formula 4T-2mg=2ma
a=?
Inaonekana a man of mass (m) hana ubavu huo.......

Wewe malizia with negative acceleration nalo ni jibu

 

[Mokaze]

What if we consider only boundary forces- 4 T upwards where t is tension force in rope and 2mg downward gravitational force.
By Newtons second law of motion F=ma: 4T-2mg=2ma . 2,unknowns(T &a) ,one equation CHECKMATE

You have to consider it by splitting the problem into two parts each one having free forces acting on.

The parts are; (1) the man standing on the platform and (2) the platform itself.

The upward forces acting on the man are the Tensions (mg/2)×2 and reaction "R" of his weight acting on the platform, the downward force acting on the man due to his weight is mg, the man moves the ropes upwards with acceleration "a" therefore the sum of all the upward forces minus all the downward forces acting on the man equals "ma" and that will be the equation #1

The equation #2 is to be found from the platform its free forces acting on and the same steps are to be followed as in the previous procedures, from the eqn #1&2 you will equate the reaction R, which will be found in both equations, to "ma" whereas you can calculate the acceleration "a".

There may be another method apart from this one, I can't rule out.

 

[Mokaze]

a=0

You are right

 

[simplemind]

We are told man pulls each rope with a force equal half his weight therefore upward force balances mans weight thus acc, is zero . The upward tension force on platform equals it weight hence its acc. is zero. System is in equilibrum.

 

[Mokaze]

We are told man pulls each rope with a force equal half his weight therefore upward force balances mans weight thus acc, is zero . The upward tension force on platform equals it weight hence its acc. is zero. System is in equilibrum.

Very simple true logic, could you please take another challenge in the same but twisted phenomenon ??

Assume both the man and the platform have masses of 60kg and 40kg respectively and the system moves upwards as the man flexes his arms and legs to produce an impetus resulting the system to move upwards with 2m/s² acceleration, 1---What will be the tension force T on each rope?? 2--what will be the reaction force the man feet exert on the platform ??, take g=10m/s².

 

[simplemind]

What if we consider only boundary forces- 4 T upwards where t is tension force in rope and 2mg downward gravitational force.
By Newtons second law of motion F=ma: 4T-2mg=2ma . 2,unknowns(T &a) ,one equation CHECKMATE

Overlooked a crucial piece of information - we know 2T=mg,hence T=mg/2. Substitute T in first equation 4(mg/2)-2mg=ma
Hence ma=2mg-2mg
ma=0;
a=0;

 

[Mokaze]

Overlooked a crucial piece of information - we know 2T=mg,hence T=mg/2. Substitute T in first equation 4(mg/2)-2mg=ma
Hence ma=2mg-2mg
ma=0;
a=0;

that's ok.✔✔

You may try If you mind tackling qns #13

 

[simplemind]

that's ok.✔✔

You may try If you mind tackling qns #13

equation of motion of man 2T+R-60g=60x2 ........(a)
equation of motion cage 2T-R-40g=40x2 ..........(b)
add a to b to eliminate R 4T-100g=200
ie 4T=1200
T=300N(tension force)
substitute T equation (b) 600-R-400=200
200-R=200
Reaction force suprisingly enough equals 0 ????? Is it

 

[pCpCp RICH THINKER]

View attachment 1752899

a= 0

 

[Teknocrat]

Tuendelee na Physics........
Ifuatayo ni kati ya maswali ya interview kujiunga na TISS...........😁


A detective is interested in finding the depth of a well where he believes a criminal has
hidden a gun. The detective drops a stone into the well, and measures the time before a
splash is heard. The total time t consists of the time it takes for the stone to hit the water
t1 and the time it takes for the sound to travel from the surface of the water to the
detectives’ ear t2. If g is acceleration due to gravity, d the depth of the well and c the
speed of sound in air then we can write:

By rearranging the formulae, find a sensible solution for the depth of the well given that
the detective counts 2.5 seconds before hearing the splash.
Take g as 9.81 m/ s2 and c as the speed of sound in air at atmospheric pressure and 20°C.

 

[Mokaze]

Tuendelee na Physics........
Ifuatayo ni kati ya maswali ya interview kujiunga na TISS...........😁


A detective is interested in finding the depth of a well where he believes a criminal has
hidden a gun. The detective drops a stone into the well, and measures the time before a
splash is heard. The total time t consists of the time it takes for the stone to hit the water
t1 and the time it takes for the sound to travel from the surface of the water to the
detectives’ ear t2. If g is acceleration due to gravity, d the depth of the well and c the
speed of sound in air then we can write:
View attachment 1755100
By rearranging the formulae, find a sensible solution for the depth of the well given that
the detective counts 2.5 seconds before hearing the splash.
Take g as 9.81 m/ s2 and c as the speed of sound in air at atmospheric pressure and 20°C.

The sensible depth "d" of the well is 28.64 metres . Taking c=343m/s.

 

[KENZY]

A man gonna passed away..😎