Physics 15 Torque (5 of 27) Tension=? Supporting a Gate

In this video I will find x=? (minimum distance) of a 100N weight of a post held up on one end against upright post (friction=0.5) with a cable.

Next video can be seen at:

[lizard]

Been stuck on this problem for ages. Thanks for the help!

[laurenbriggs]

I got this question in an assignment and had NO idea how to answer it so BLESS YOU SIR

[leraivanets]

I’ve been stuck on this question for solid 2 hours, I study in the UK so we call those forces not torques but moments. After searching the whole internet and almost giving up, your video was just like a miracle! I finally understood every single bit of explanation and what I was doing wrong. Now it’s just left to interpret the solution to UK standards and the values we were given, and I’m eventually good to go 🎉

[NotLegato]

i want to clear up the confusion about the torque the other comment had.

using the perpendicular method, (1m)(mg) becomes -mg.

if we have a rectangle of 2m * 1m, that means the length of the hypotenuse is sqrt(5), and the midway point is sqrt(5)/2 away from the pivot.
the angle between the weight and the pivot can be found with theta = tan^-1(1/0.5) = 63.4 degrees. (63.43494 for accuracy later on)

now if we take the CROSS-PRODUCT instead of the perpendicular torque, we get r x F = sqrt(5)/2 * mg * sin(tan^-1(2)), which is exactly equal to 1, so 1*mg => -mg.

mostly i wanted to say this, because i got very confused about it myself, as well, and spent a good chunk of time figuring out, what i was doing wrong. mostly, i was working with the assumption that the angle would be 30 instead of the true 26.57.

[Dina-heuc]

i am confused as to why when calculating torque of B, FAy isnt used but when calculating torque of A, FBx is used? EDIT: nevermind, I understood it now, its because there is no perpendicular line from the force to pivot B

[gauribhardwaj]

This helped so much! Thank you for explaining it so well!!

[RawdaAHafez]

Can't we just resolve the tension into its x and y components since we have the angle 30, and deal with the y component which is Tsin(30) instead of all that?

[KabeloMashego-zvro]

could we not have used sum of forces in the x direction to find Fbx, which is equal to tcos30?

[emilyspeckhals]

this video is a lifesaver-- thank you so much!!

[everything_strength]

When calculating the Torque around B you wrote that the gate = mg is contributing a torque = -mg(1). Why can you do this? Isn't the distance between B and the center of mass found by using the Pythagorean Theorem ?

i.e.

(1^2) + (0.5^2) = hyp^2 ----> hyp = 1.118.. Then, the perpendicular component is mgsin(26.565)

[SoheeChoi-dn]

when you find the tension can you also you use the method of TLcos(theta)?

[lightvsdark]

So did you figure out how to separate Fay and Fby? Since Fa and Fb are both acting in the hinges in a contact force couldn't they be combined into a normal force? And if they are a normal force then couldn't the y components just be divided evenly between the 293N? Or maybe move the pivot point to the point of attachment of the cable to the gate so that the Fay and Fby forces are not parallel to the axis of rotation and then solve with rotational equilibrium statements like we did for the others? Love your vids. This cliff hanger is killing me hahahah! Hope we can solve for the other two force vectors!

[asad]

At around 8:00 why did you use d2 instead of d?