Man Moving Crate: Physics Homework Help

  • Thread starter Discalculiate
  • Start date
In summary, the conversation is about a student seeking help with practice questions for a midterm. The questions involve calculating force and friction for a 100 kg crate being pulled to the right by a person with a force of 150 N. The student is unsure of the specific equations needed and is asking for clarification. The conversation also touches on the concept of coefficients of static and dynamic friction.
  • #1
Discalculiate
4
0
My teacher gave us practice questions for the midterm. Can't for the life of me remember how to do this. Any hints would be greatly appreciated. I'm not just trying to bum answers off you guys; I really don't know where to start.

Homework Statement



Чs = .2 (the Ч is backwards - can't find a closer symbol, but you know what I mean)
Чk = .1
A man is pulling a 100 kg crate to the right. Fman = 150 N
a) What is the force? [er, not entirely sure what he means by this?]
b) Fman to get the crate moving?
c) Fman to move crate at constant speed?

Homework Equations



Unknown

The Attempt at a Solution



heh.
 
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  • #2
You must show some work to get help here. What is the general equation relating the horizontal force on an object like that crate, the weight of the crate, and the coefficients of static and dynamic friction [tex]\mu_s[/tex] , [tex]\mu_d[/tex] ?
 
  • #3
Well I know fsmax = Чs(N)(9.8)

So fmax = (.2)(150)(9.8)? which would be 294 N

Is that what he wants for part a?
 

FAQ: Man Moving Crate: Physics Homework Help

1. What is the concept behind "Man Moving Crate" in physics?

The concept behind "Man Moving Crate" in physics is that of work and energy. The man applies a force to the crate, causing it to move a certain distance. This requires the man to do work, which is the product of the applied force and the distance moved. The energy used by the man in moving the crate is transferred to the crate, giving it kinetic energy.

2. How does the weight of the crate affect the amount of work done by the man?

The weight of the crate affects the amount of work done by the man as it determines the force required to move the crate. The heavier the crate, the more force the man has to apply to move it a certain distance, thereby doing more work. This relationship can be expressed by the equation W = Fd, where W is the work done, F is the applied force, and d is the distance moved.

3. What role does friction play in the movement of the crate?

Friction plays a significant role in the movement of the crate as it opposes the motion of the crate. This means that the man has to apply a greater force to overcome the frictional force and move the crate. Friction also causes the crate to lose some of its kinetic energy as heat, thereby reducing its speed and making it more difficult to move.

4. How does the angle of incline affect the work done by the man?

The angle of incline affects the work done by the man as it changes the distance over which the crate is moved. For example, if the crate is moved up an incline, the distance traveled is longer than if it was moved horizontally. This means that the man has to apply a smaller force to move the crate up the incline, but over a longer distance, resulting in the same amount of work being done.

5. Can you calculate the amount of work done by the man in moving the crate?

Yes, the amount of work done by the man in moving the crate can be calculated using the equation W = Fd, where W is the work done, F is the applied force, and d is the distance moved. Additionally, if the angle of incline and the weight of the crate are known, the work can be calculated using the formula W = mgh, where m is the mass of the crate, g is the acceleration due to gravity, and h is the vertical height the crate is moved.

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