Achieving Horizontal Motion of a Crate: A Physics Challenge

In summary, the person wants to cause a 200N crate to start sliding towards the right by exerting a force on it through a rope. The horizontal component of the force must be 0.35 times the normal force exerted by the floor. The angle between the rope and the horizontal is 20 degrees. The equations Ff = μFN and Fapp = 70N are used to solve for the force needed to move the crate, but the book suggests a different answer. In another example, the rope is anchored at a point and pulled upwards at a 10 degree angle, causing the angle between the rope and the ground to change. Equation Ff = μFN is still used to solve for the force needed to
  • #1
Melawrghk
145
0

Homework Statement


The person wants to cause the 200N crate to start sliding toward the right. To achieve this, the horizontal component of the force exerted on the crate by the rope must equal 0.35 times the normal force exerted on the crate by the floor.

[[[[[[[[[ +
[...[ +
[...[+
[[[[[[[[[

The box is the crate and the pluses represent the rope the guy is pulling. It makes an angle of 20 degrees with the horizontal.

Homework Equations


Ff=[tex]\mu[/tex]FN

The Attempt at a Solution


So I figured that 0.35 is the [tex]\mu[/tex] for friction. Since its a leveled surface, FN will equal Fg and thus they cancel out. The horizontal component of the Fapp has to equal Ff because he's trying to move it, not accelerate it.

From this, I got Fapp to equal 70N. The angle is 20 degrees, so the Fapp is 74.5N (got by: 70N/cos(20)). But the back of the book is telling me it should be 66N. I don't know how I'd get that. But then again, its Engineering Statics, so maybe there is some trick? Please help?
 
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  • #2
Well you have a problem here. You say the only y forces are normal force and gravity.

However there is a y component of the man pulling the crate.
You need to form two simultaneous equations.
You do NOT know the mass of the crate.

So do this: let x = the force he is pulling with (hypotenuse).

[tex] 200 = mg + xsin(20) [/tex]
[tex] xcos(20) = 0.35mg [/tex]

Now explain to me why these equations are as they are.
You will get the right answer.
 
  • #3
Why do we not know the mass of the crate? We know the weight, which is m*9.81?

Okay.. I'll try:
200=mg+xsin(20).. In this one 200 is the upward force created by two things - the man lifting the object (xsin(20) is the y-component of him lifting it) and mg is simply opposite the force the ground is exerting on the object upwards. They're added because they are the forces acting up.

xcos(20)=0.35mg, well that's a given. The horizontal component of the applied force is 0.35 of the natural force.

I can picture this in a FBD, I think. Thanks!
 
  • #4
Melawrghk said:
Why do we not know the mass of the crate? We know the weight, which is m*9.81?[...]

we know the net force in the negative y direction, 200N. However, this is mg + xsin(20) because there is mg pushing down and a small component of the man pulling up. If he was only pulling horizontally, it would be m*9.81.
We can however, using those simultaneous equations solve for the mass of the crate. When I said we do not know the mass, I meant, we have not been given the mass in a simple form.

Your explanation is enirely right.
 
  • #5
What about this though?
box.jpg


Its the same box, but I don't know how to set this up. Isn't the angle changing as he's pulling the rope upwards?
 
  • #6
yes, in that example it is. However, in the first example that you posed; he is pulling parallel to the angle of the chain. Therefore the angle of the chain is NOT changing but there is still an X and Y component of force.
 
  • #7
Yeah, I know in the first one it wasn't changing. How would I solve this one though?
 
  • #8
What are you given, and what is the question asking?
 
  • #9
It's the same question. Only now he anchored the cable to a point (the blue circle on the right) and is pulling the rope upwards. The 10 degrees refers to the angle the rope makes with the ground at the point where it's anchored. Same box (200N).
 
  • #10
And the rope attached to the crate is horizontal?
 
  • #11
It's not parallel to the ground in the book... So I guess not.
 

Related to Achieving Horizontal Motion of a Crate: A Physics Challenge

1. How do you achieve horizontal motion of a crate?

To achieve horizontal motion of a crate, an external force must be applied to the crate in the direction of motion. This force can be provided by pushing or pulling the crate, or by using a machine such as a conveyor belt.

2. What is the physics behind achieving horizontal motion of a crate?

The physics behind achieving horizontal motion of a crate involves Newton's laws of motion. The first law states that an object at rest will remain at rest, and an object in motion will continue in motion at a constant velocity, unless acted upon by an external force. The second law states that the acceleration of an object is directly proportional to the net force applied to it and inversely proportional to its mass. This means that a larger force is required to move a heavier crate, and a smaller force is needed to move a lighter crate. The third law states that for every action, there is an equal and opposite reaction. This means that when a force is applied to the crate, the crate will exert an equal force in the opposite direction.

3. What factors affect the horizontal motion of a crate?

The factors that affect the horizontal motion of a crate include the magnitude and direction of the applied force, the mass of the crate, and any external factors such as friction or air resistance. The surface on which the crate is moving and the angle at which the force is applied can also affect its motion.

4. How can you increase the horizontal motion of a crate?

To increase the horizontal motion of a crate, you can increase the magnitude of the force applied to it, decrease its mass, or reduce any external factors that may be slowing it down. You can also change the angle at which the force is applied to optimize its motion.

5. What are some examples of real-world applications of achieving horizontal motion of a crate?

Some examples of real-world applications of achieving horizontal motion of a crate include loading and unloading cargo from ships or trucks, moving packages along a conveyor belt in a warehouse, and pushing a shopping cart in a store. Other examples include using a forklift to move heavy crates in a factory, or sliding a hockey puck across the ice during a game.

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