Recent content by Specter

Sorry but I am still confused. ##F_{f_k} - F_{f_s}= \mu_k mg \cos \theta - \mu_s mg \cos \theta## ##F_{f_k} - F_{f_s}=mg\cos \theta(\mu_k-\mu_s)## ##F_{f_k} - F_{f_s}=\mu_k-\mu_s## If I factor ##mg\cos \theta## out of the right hand side, I am left with just ##\mu_k-\mu_s## and honestly, I am...

So when subtracting the two equations: ##\mu_smg\cos (\theta)-\mu_kmg\cos (\theta)## Wont I just be left with ##\mu_s-\mu_k##?

How would I subtract those algebraically? ##F_f=\mu_smg\cos (\theta)-\mu_kmg\cos (\theta)## If I use the same mass for each equation wouldn't I get the same result as above?

When solving for the force of friction in each of the equations above, does it matter which mass I use (10 kg or 20 kg)? This is what I've done. I really hope this is correct. ##F_{f_s}=\mu_smg\cos (\theta)## ##F_{f_s}=(0.4)98\cos21.8## ##F_{f_s}=36.4 N## ##F_{f_k}=\mu_kmg\cos (\theta)##...

Alright. So I have the normal force which is ##F_N=mg\cos \theta##. To find the force of kinetic friction I would do ##F_{f_k}=\mu_kmg\cos \theta## To find the force of static friction I would do ##F_{f_s}=\mu_smg\cos \theta## Is this the correct way to figure that out?

##F_N=(10 kg)(9.8m/s^2)\cos21.8## ##=91 N## I've got this answer before, am I still solving incorrectly? I also have two masses, 10 kg and 20 kg, so do I find the normal force for both?

You said that the normal component of the gravitational force plus the normal component of the contact force will total to give you the net force in the normal direction. I now see that the normal component of the contact force is ##F_N## because ##F_N## is perpendicular to ##F_s##. I should...

Yeah at that point I was just guessing. All I need to do is add the components of mg?So ##F_g## is the force of gravity. ##mg\sin \theta## would be the normal component of the gravitational force. ##mg\cos \theta## is the normal component of the contact force. So...

Ahhh I am so lost. So ##mg\cos \theta## does not have a direction because it is not a force, but instead just a component of ##mg##. Would ##F_N=F_g## then? I don't get it. As for the change in frictional force I need to find the normal force before doing that. Once I have found ##F_N##...

##F_{net}## is a sum of ##F_N## and ##\displaystyle F_{g_y}##, but ##F_N=0## because there is no acceleration along the y-axis. Is that correct? I used ##mg\cos \theta## because ##F_{g_y}=mg\cos \theta## . I don't really understand this much but I'm trying my best to find problems in my text and...

So how about this? ##F_N## and ##mg\cos (\theta)## both act in the same direction so this is right? \begin{array}{l} F_{net}=F_N-F_g \\ F_{net}=F_N-mg \\ 0 N=F_N-mg \cos \theta \\ F_N=mg\cos \theta \\ F_N=(10)(9.8)cos21.8 \\ F_N=90.99 \\ F_N=91 N \end{array} How do I find the change in...

The change in the frictional force would be 0.1. The motion is only occurring along the x-axis, right? So the net force in the y-direction would be 0? Which expression should I use? I feel like the second one I wrote is the correct one because I tried solving with the first one and it works out...

In my text this is how a similar example had solved the problem. Have I done this correctly?The normal force of box a (10 kg) is 91 N The normal force of box b (20 kg) is 182 N Box A: ##F_f=\mu F_N## ##F_f=(0.1)(90.00)## ##F_f=9.01 N## Box B: ##F_f=\mu F_N## ##F_f=(0.1)(182.00)## ##F_f=18.2 N##

Alright. This would be the equation to calculate the normal force. I would have to do this for each box because of the different masses right? 10kg and 20kg. Hopefully this is correct. I just have to plug everything in. \begin{array}{l} F_{net}=F_N-F_g \\ F_{net}=F_N-mg \\ 0 N=F_N-mg \cos...

Before it broke loose the acceleration was ##a=0.0m/s^2## After it broke loose, the friction changed from ##\mu_s=0.4## to ##\mu_k=0.3##. I guess this would result in a change of 0.1 in friction? I haven't had to do that yet. How can I find how the acceleration will change from the change in...