# Two men, a crate, and a set of stairs

• TFM
In summary: Homework Equations:The Attempt at a SolutionThe forces applied to the box have to sum to g*(200kg) and they also have to create zero net torque around the center of mass of the box. Otherwise the box won't stay at a constant angle. That will give you two equations in two unknowns. That's where to start.I think I must have my torque equation wrong:I haveF1+F2=1960 (<-200g)My Torque calucluations:F1D = F2DF1*cos(246.8)*1.346 = F2*cos(68.2)*1.
TFM
[SOLVED] Two men, a crate, and a set of stairs

## Homework Statement

Two men are carrying acrate up some stairs, at an angle of 45 degrees. the crate weighs 200kg, and the box is 1.25m long by 0.5m high. The force each person applies is verticle. I need to work out the force on each person, but I am not sure where to start. Any ideas?

## The Attempt at a Solution

Last edited:
The forces applied to the box have to sum to g*(200kg) and they also have to create zero net torque around the center of mass of the box. Otherwise the box won't stay at a constant angle. That will give you two equations in two unknowns. That's where to start.

I think I must have my torque equation wrong:

I have
F1+F2=1960 (<-200g)

My Torque calucluations:
F1D = F2D
F1*cos(246.8)*1.346 = F2*cos(68.2)*1.346

which I rearrange to get:

-772.126F1 = 727.881F2

F1=1960-F2

-772.126(1960-F2) = 727.881F2
-1513367+772.126F6 = 727.881F2
-1513367 = -44.2F2
F2=34201

Which is definitely wrong!

The Actual Answers are F1 = 590, F2 = 1370

Any ideas?

TFM said:
I think I must have my torque equation wrong:

I have
F1+F2=1960 (<-200g)
OK.

My Torque calucluations:
F1D = F2D
F1*cos(246.8)*1.346 = F2*cos(68.2)*1.346
Not sure what you're doing here. What pivot point are you using to calculate torque? Where did you get a distance of 1.346? (That's the diagonal across the side of the crate.)

Hint: Where's the center of mass of the crate?

I was taking the torques from the centre of mass, which I asumed was at the centre of the rectangle, since there was no mention of it being non-uniform

TFM

OK. So explain how you got your angles and distances in your torque equation. (A diagram showing the forces would be helpful.)

The Length of the diagonal was calculated using Pythgoras (x^2+y^2 = Z^2) although the length from the pivot to the force is actually half that length (error)

The angles were calculated using the dimensions of the box to find the inner angle

F1(1960*cos23.3)*0.67 = F2(1960*cos246.8)*0.67
-519F1 = 1212.675F2
F1 = 1960-F2

-519(1960-F2)=1212.675F2
-1018717 +519F2 = 1212.675F2

Which works out as F2 = -1470, which is near the value required but 1=hundred out, excluding the fact that it is a negative.

TFM

#### Attachments

• Crate Diagram.bmp
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TFM said:
The Length of the diagonal was calculated using Pythgoras (x^2+y^2 = Z^2) although the length from the pivot to the force is actually half that length (error)
OK.

The angles were calculated using the dimensions of the box to find the inner angle

F1(1960*cos23.3)*0.67 = F2(1960*cos246.8)*0.67
Why is there a 1960 in there? Why are you using cosines? Since you are setting clockwise torques = counterclockwise torque, you can forget about the signs--just worry about magnitudes.

the 1960 and cosine are in thereto determine the torque, since:

Torque = Force * Perp. Distance, the forces and distnbace a=ren't perpendicular, so I used 1960*cos(angle), to calculate the perpendicular force for each Torque.

TFM

Since the forces are all vertical, the perpendicular distance would be horizontal. You'd need the horizontal component of the distance and thus the angle it makes with the horizontal to use cosine. (Since you have the angle with the vertical, you can use sine.)

Also: 1960 is the total force; it shouldn't appear in the torque equation.

I now have:

F1*(sin246.8)0.63 = F2*(sin23.2)0.63

for the Torque equation, which rearrnges to:

-0.61871F1 =0.26518F2

F1 = (1960 - F2)

-0.61871(1960 - F2) = 0.26518F2
-1212.68 + 0.61871F2 = 0.26518F2
-1212.68 = -0.35353F2,

Which gives F2 to be 3430!?

TFM

TFM said:
I now have:

F1*(sin246.8)0.63 = F2*(sin23.2)0.63
Since all you care about are the magnitudes (you already know which is clockwise, etc.), use angles between 0 and 180. Or just ignore the signs of your sines.

for the Torque equation, which rearrnges to:

-0.61871F1 =0.26518F2
Lose the negative sign. (Realize that a negative sign makes no sense!)

F2 = 1212.675/0.883892 = 1371!

F1 = 1960 - F2

Thanks Doc Al, great job

TFM

## 1. What is the significance of two men, a crate, and a set of stairs in scientific experiments?

There is no specific significance to these objects in scientific experiments. They could be used as props or tools in certain experiments, but they do not have any inherent scientific significance on their own.

## 2. Can these objects be used to demonstrate principles of physics?

Yes, depending on the specific experiment, these objects could potentially be used to demonstrate principles of physics such as gravity, force, and friction.

## 3. How do scientists ensure safety when using a crate and stairs in experiments?

Scientists follow strict safety protocols and guidelines when conducting experiments, including proper handling and use of equipment, safety gear, and precautions to prevent accidents or injuries.

## 4. Are there any famous experiments that involve two men, a crate, and a set of stairs?

There are many famous experiments that use these objects, such as the Stanford Prison Experiment and the Milgram Experiment. However, these objects are not the focus or key component of the experiments.

## 5. Can these objects be used in psychological experiments?

Yes, these objects could potentially be used in psychological experiments as part of the experimental design or as a tool for data collection. However, it would depend on the specific experiment and its objectives.

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