physics Physics help?

[funkeemunkee]

I have this take home assignment for physics and I am super stuck on these 3 problems... see photo 

[Dissitesuxba11s]

Basic formula to remember: F=m*a

Units:

Force is in Newtons (N)

Mass is in kilograms (kg)

Acceleration is in meters per second squared (m/s²) 

 

#1

Calculate the net force (F_Net) by multiplying mass (m) by the acceleration (a). Note that acceleration is negative. You can solve for force 1 (F₁) by summing all the left and right forces and equaling it to the net force (F_Net).

 

#2

No acceleration is present so the sum of forces is 0. This means the forces on the right side are equal to the ones on the left side.

 

#3

One Gram (g) is equal to 0.001 kg. Normal force (F_N) is the mass of the box multiplied by gravity. Net force (F_Net) is the sum of all the forces in one directional plane, in this case the X-axis. The acceleration (a) of the box is net force (F_Net) divided by the mass (m).

Edited December 2, 2017 by Dissitesuxba11s
Corrected the axis

[funkeemunkee]

2 minutes ago, Dissitesuxba11s said:

Basic formula to remember: F=m*a

Units:

Force is in Newtons (N)

Mass is in kilograms (kg)

Acceleration is in meters per second squared (m/s²) 

 

#1

Calculate the net force (F_Net) by multiplying mass (m) by the acceleration (a). Note that acceleration is negative. You can solve for force 1 (F₁) by summing all the left and right forces and equaling it to the net force (F_Net).

 

#2

No acceleration is present so the sum of forces is 0. This means the forces on the right side are equal to the ones on the left side.

 

#3

One Gram (g) is equal to 0.001 kg. Normal force (F_N) is the mass of the box multiplied by gravity. Net force (F_Net) is the sum of all the forces in one directional plane, in this case the Y-axis. The acceleration (a) of the box is net force (F_Net) divided by the mass (m).

Well **** didn't expect anyone would help  thanks!

[Dissitesuxba11s]

Just now, funkeemunkee said:

Well **** didn't expect anyone would help  thanks!

No problem, hope it made sense.

[funkeemunkee]

1 minute ago, Dissitesuxba11s said:

No problem, hope it made sense.

It did!

[funkeemunkee]

1 hour ago, Dissitesuxba11s said:

No problem, hope it made sense.

I came back to it and -18,560N doesn't seem correct for f1 is it?

[funkeemunkee]

32 minutes ago, funkeemunkee said:

I came back to it and -18,560N doesn't seem correct for f1 is it?

Also this last thing 

[Dissitesuxba11s]

3 hours ago, funkeemunkee said:

.

-18,560N is not the right answer. Remember your signs change depending on the direction.

 

Abigail Problem:

Rolling force is equal to normal force times rolling coefficient. Remember that normal force is mass time gravity.

The force that the car exerts on the road (F_car) to move is essentially a positive force while the rolling force (F_roll) is an opposing force going the other direction. Adding these two forces (F_car & F_roll) will give you the net force, which you can calculate using the basic force equation.

 

I'll look at the rest in the morning. Why is it number 11 then 9?

 

 

 

[Dissitesuxba11s]

14 hours ago, funkeemunkee said:

.

Before I show you the fundamentals of how to solve 11 and 9, do you know the basic kinematic equations? I don't want to show you something that your teacher hasn't taught you since that would be suspicious.

[funkeemunkee]

3 hours ago, Dissitesuxba11s said:

Before I show you the fundamentals of how to solve 11 and 9, do you know the basic kinematic equations? I don't want to show you something that your teacher hasn't taught you since that would be suspicious.

Yes he has taught that and the numbers confuse me aswell

[funkeemunkee]

5 hours ago, funkeemunkee said:

Yes he has taught that and the numbers confuse me aswell

Last ones  assume all equations are fair game because they are

[funkeemunkee]

This is why you bring your notebook full of notes HOME with you... I'm bad

[Dissitesuxba11s]

20 hours ago, funkeemunkee said:

.

11)

To determine if the box will move, you would have to calculate whether the force applied is greater that the force of static friction (F_Stat). You cannot simply use the force that is given in the problem due to the fact that it is at an angle. Imagine a right triangle and the force applied is along the hypotenuse. To calculate the forces applied in the X and Y directions you would have to multiply the initial force by the sine or cosine of the angle, respectively.

 

To calculate your friction forces for both static and sliding, you would need to calculate the normal force (mass times gravity) and multiply it by the respective coefficient of friction. Now that you have all prevailing forces, you simply have to compare the force applied in the Y direction to the force of static friction. If the force force applied is greater or equal to the force of static friction, then the box will move.

 

Calculating the time it take for the box to travel the length of the room requires you to use the kinetic equation:

d = v_i t + 0.5 a t²

Where:

d= distance

v_i= initial velocity

t= time

a= acceleration

 

First you would have to calculate acceleration. Use F=ma, but use the force in the Y-direction. Now that you have acceleration, you just have to solve for time (t), remembering that since is started from a complete stop, initial velocity (v_i) is zero.

 

[Dissitesuxba11s]

21 hours ago, funkeemunkee said:

.

9)

A. First calculate, your friction forces on mass A, F_Stat and F_Slide . Now calculate your force applied by masses C and B, using the provided number for gravity. To determine if mass A will move from rest, you would have to see if the sum of the external forces (F_Ext) is equal or greater to the force of static friction (F_Stat). Remember that masses C and B are going in the opposite direction.

 

B. With the previously calculated sum of the external forces (F_Ext) and force of sliding friction (F_Slide), you know have to determine the net force (F_Net) applied on mass A. Remember what direction F_Ext is going to, and that F_Slide will be going against it. Now that you have F_Net calculated, you simply have to use F=ma to calculate acceleration.

[Dissitesuxba11s]

22 hours ago, funkeemunkee said:

.

14)

Source: http://www.dummies.com/education/science/physics/calculating-the-force-needed-to-move-an-object-up-a-slope/

 

I'm using this image since I didn't want to sketch it myself; just imagine it's the same orientation as the problem. First, you would need to calculate your force due to sliding (F_Slide) by multiplying the coefficient of sliding friction by the normal force at a cosine of 60 degrees. F_Slide will be opposed ONLY by the "force" of gravity (F_Fall), which is calculated by the normal force by sine 60 degrees. 

 

A. To determine if the box will slide off or not, you would need multiple kinematic equations.

 

Source: http://www.physicsclassroom.com/class/1DKin/Lesson-6/Kinematic-Equations

 

First, you would need to calculate the velocity of the box at the end of the incline (v_f) using the top right equation, with the acceleration (a) coming from F_Fall. Remember that since this acceleration is due to gravity it would be negative. If your value for v_f is positive, then the answer to part A is "Yes, the box will slide off." If your value for v_f is negative, then your answer would be "no."

 

B. IF YES:

Use the last formula below to calculate the horizontal range, using your calculated calculated value for v_f as v₀ .

 

Source: http://formulas.tutorvista.com/physics/trajectory-formula.html#

 

IF NO:

Use the bottom left kinematic equation to solve for your time, you value for v_f as zero. Now use the bottom right kinematic equation to solve for distance still using zero final velocity.

 

This should be it. Hopefully you see this in time, and that it helped you understand these problems.

[funkeemunkee]

58 minutes ago, Dissitesuxba11s said:

14)

Source: http://www.dummies.com/education/science/physics/calculating-the-force-needed-to-move-an-object-up-a-slope/

 

I'm using this image since I didn't want to sketch it myself; just imagine it's the same orientation as the problem. First, you would need to calculate your force due to sliding (F_Slide) by multiplying the coefficient of sliding friction by the normal force at a cosine of 60 degrees. F_Slide will be opposed ONLY by the "force" of gravity (F_Fall), which is calculated by the normal force by sine 60 degrees. 

 

A. To determine if the box will slide off or not, you would need multiple kinematic equations.

 

Source: http://www.physicsclassroom.com/class/1DKin/Lesson-6/Kinematic-Equations

 

First, you would need to calculate the velocity of the box at the end of the incline (v_f) using the top right equation, with the acceleration (a) coming from F_Fall. Remember that since this acceleration is due to gravity it would be negative. If your value for v_f is positive, then the answer to part A is "Yes, the box will slide off." If your value for v_f is negative, then your answer would be "no."

 

B. IF YES:

Use the last formula below to calculate the horizontal range, using your calculated calculated value for v_f as v₀ .

 

Source: http://formulas.tutorvista.com/physics/trajectory-formula.html#

 

IF NO:

Use the bottom left kinematic equation to solve for your time, you value for v_f as zero. Now use the bottom right kinematic equation to solve for distance still using zero final velocity.

 

This should be it. Hopefully you see this in time, and that it helped you understand these problems.

You are so awesome thank you! Is there any way to gift a floatplane or contributor rank?

[funkeemunkee]

You are so awesome thank you! Is there any way to gift a floatplane or contributor rank?

[Dissitesuxba11s]

17 minutes ago, funkeemunkee said:

You are so awesome thank you! Is there any way to gift a floatplane or contributor rank?

I don't think so, but either way it's alright, I'm just glad to help. Plus, it was a good refresher of physics.

[funkeemunkee]

2 minutes ago, Dissitesuxba11s said:

I don't think so, but either way it's alright, I'm just glad to help. Plus, it was a good refresher of physics.

Either way, thank you for the help

[Dissitesuxba11s]

8 hours ago, funkeemunkee said:

Either way, thank you for the help

No problem 

[VerticalDiscussions]

4 hours ago, Dissitesuxba11s said:

No problem 

Lol, man... this is some hardcore thing right here (for me it is, insanely, alien)!

 

I clicked this topic and thought:

 

I read it and, afterwards, was like:

 

 

Ahhhhhh, halp!!

[funkeemunkee]

7 minutes ago, VerticalDiscussions said:

Lol, man... this is some hardcore thing right here (for me it is, insanely, alien)!

 

I clicked this topic and thought:

 

I read it and, afterwards, was like:

 

 

Ahhhhhh, halp!!

That would be AP physics for ya!

[Teddy07]

no offense but you are supposed to it yourself and not others do your complete homework

[Dissitesuxba11s]

Just now, Teddy07 said:

no offense but you are supposed to it yourself and not others do your complete homework

Agreed, which is why I didn't necessarily do the work for OP. I explained it in a way that it focuses on the fundamentals of the topic and the problem solving method. This way OP can apply it to other problems in the future.

[funkeemunkee]

1 hour ago, Dissitesuxba11s said:

Agreed, which is why I didn't necessarily do the work for OP. I explained it in a way that it focuses on the fundamentals of the topic and the problem solving method. This way OP can apply it to other problems in the future.

Which it did help and I WAS allowed to use other people to help me that's why I asked