What is the Work and Normal Force in a Block Pushed Up a Wall?

[Warmacblu]

Work = F*d*cosθ, where θ is the angle between the force and the displacement. Note that F*cosθ can be viewed as the component of force in the direction of the displacement. If there's no component of force in the direction of the displacement, then that force does no work.

This might help a bit: http://hyperphysics.phy-astr.gsu.edu/hbase/work2.html#wdef"

I love HyperPhysics, I have been referencing that site since day one.

I got the correct answer for the first part ... 142.99. Now onto the third part.

I am asked to find the magnitude of the normal force between the block and the wall. The normal force is being exerted by the wall to counter the force I put on the block.

The force I put on the block is what I just solved ... Fdcos(theta) and the work done by the normal force is ...

I can't seem to come up with a definite answer but I can take a stab at it.

I believe the normal force would be the opposite of the force put on the block, so -Fdsin(theta)?

I am having a hard time understand the normal force for this problem.

Any help is appreciated.

I just thought about this. The normal force is the same force exerted by the first force from part 1 so I have to convert that number into N to get my answer. But I don't know how that would work if the force is at an angle.

 

[Doc Al]

Part C just asks for the magnitude of the normal force. Go back to your original equation for the horizontal forces, which will tell you N in terms of F (which you already know now). (Make sure you use the proper angle and trig function.) This part has nothing to do with work.

 

[Warmacblu]

Part C just asks for the magnitude of the normal force. Go back to your original equation for the horizontal forces, which will tell you N in terms of F (which you already know now). (Make sure you use the proper angle and trig function.) This part has nothing to do with work.

Alright, my original equation is:

N - Fdsin27

So N = Fdsin27

I know F = 142.99
I know d = 3m

Now about the angle and trig function ... I believe the trig function will be cos because the angle is 27 degrees above the horizontal and the angle will still be 27.

 

[Doc Al]

Alright, my original equation is:

N - Fdsin27

So N = Fdsin27

This is incorrect:
(1) You have the wrong trig function
(2) You are mixing work with force (work is irrelevant here)

Now about the angle and trig function ... I believe the trig function will be cos because the angle is 27 degrees above the horizontal and the angle will still be 27.

This is true, so correct your equation.

 

[Warmacblu]

This is incorrect:
(1) You have the wrong trig function
(2) You are mixing work with force (work is irrelevant here)


This is true, so correct your equation.

Since cos and 27 degrees is correct, I believe N = Fcos27. However the N force is in the negative direction. I did try N = Fcos27 in the positive direction and that answer was incorrect. I just want to make sure that the answer has to be negative before I use any more tries.

Thanks.

 

[Doc Al]

Since cos and 27 degrees is correct, I believe N = Fcos27. However the N force is in the negative direction. I did try N = Fcos27 in the positive direction and that answer was incorrect.

I don't see why that would be incorrect:
ΣF_x = 0
Fcos27 - N = 0
So: N = Fcos27

What value of F are you using? Be careful not to confuse force with work!

I just want to make sure that the answer has to be negative before I use any more tries.

They want the magnitude, so your answer must be positive.

 

[Warmacblu]

I don't see why that would be incorrect:
ΣF_x = 0
Fcos27 - N = 0
So: N = Fcos27

What value of F are you using? Be careful not to confuse force with work!

They want the magnitude, so your answer must be positive.

I used the value I got from the first part which is work. I guess I have to take that work and divide it by 3 to get the force?

Edit - I have to solve the force from my first equation in the y-direction. I will report back in a bit.

 

[Warmacblu]

I solved it.

I took my answer from when I solved the force for the previous part ... 104.99 and multiplied that by cos27.

Thanks for all the help with this problem, I appreciate it.

 

[Doc Al]

You already solved for the force F. Use the value you found earlier.

 

[Warmacblu]

I guess you posted a few seconds after. I managed to figure out that is the force I am supposed to use.

Thanks again.