1. ## statics

can someone help me with question 7b on this. its the marking scheme of the question page 16.

https://www.examinations.ie/tmp/1502109748_6487413.pdf

I can understand it until it says Q=T
where does this come from.

thanks

2. ## Re: statics

recheck your link ... I get the exam page, not a pdf file.

5. ## Re: statics

can you see that? sorry for the first post

6. ## Re: statics

Originally Posted by edwardkiely
can someone help me with question 7b on this. its the marking scheme of the question page 16.

I can understand it until it says Q=T
where does this come from.
Q is the weight of the load directed vertically downward

T is the tension anywhere in the rope ... T is the force pulling vertically upward on load Q.

The load is in static equilibrium $\implies$ Q = T

7. ## Re: statics

Originally Posted by skeeter
Q is the weight of the load directed vertically downward

T is the tension anywhere in the rope ... T is the force pulling vertically upward on load Q.

The load is in static equilibrium $\implies$ Q = T

in these type of questions are you not supposed to let all forces up = forces down on the whole system? what about all the other forces like R and W

are they only focusing on a certain part of the whole system?

8. ## Re: statics

Yes, since this is "statics", there is no motion, all forces are balanced. One method of solving this is to break all forces into vertical and horizontal components, then set the sum all "upward" components equal to the sum of all "downward" components and set the sun of all "rightward" components equal to the sum of all "leftward components".

Another, equivalent, method is to make all "upward" and "rightward" components positive, all "downward" and "leftward" components negative and set the signed sum of all components equal to 0.

9. ## Re: statics

ok thanks . so there are methods to solve it.
but how come in the marking scheme they are only letting Q=T. I can not understand how they are just focusing on two forces of the whole system and just letting them equal. there other forces beside the Q and T keeping it in equilbrium. I am confused on how they decided they were just going to pick the two forces Q and T and how they decided what section of the whole system to look at. i am assuming they are only concerned about the forces on the string when they form the equation Q=T and not about the rest of the system.

they did not even resolve the tension of the string.

10. ## Re: statics

When one analyzes a system in equilibrium, it is common practice to isolate each component of that system individually since each are separately in a state of static equilibrium.

In this system, there are three individual "components" ...

(1) the hanging mass with weight $Q$

(2) the upper "peg" at point $c$

(3) the rod touching the surface (ground) at end $a$ with length $h$ and weight $W$.

... an isolation sketch of the forces acting on each individual component is attached.

Starting with component (3), the rod has both translational and rotational equilibrium. Using torques about point $a$ ...

$\displaystyle \sum \tau_a = 0 \implies T \cdot h = W\cos{\theta} \cdot \dfrac{h}{2} \implies T = \dfrac{W\cos{\theta}}{2}$

since $\theta = 30^\circ$, $T = \dfrac{W\sqrt{3}}{4}$ ...

Going to component (1), the hanging mass with weight $Q$ is in equilibrium in the vertical direction $\implies Q=T \implies Q = \dfrac{W\sqrt{3}}{4}$

The question did not ask for them, but let's look at the reaction forces on the rod at point $a$ ...

$\displaystyle \sum F_y = 0 \implies T\cos{\theta} + R_a = W \implies R_a = W - T\cos{\theta} = W - \dfrac{W\cos^2{\theta}}{2} = W\left(1 - \dfrac{3}{8}\right) = \dfrac{5W}{8}$

An aside ... I disagree that the force of friction $f$ acting on the rod equals $\mu \cdot R_a$. That would be true only if the force of static friction were at a maximum. A correct relationship would be $f \le \mu \cdot R_a$. However, $f$ can be determined by setting up an equilibrium condition for the rod in the horizontal direction ...

$\displaystyle \sum F_x = 0 \implies f = T\sin{\theta} = \dfrac{W\sqrt{3}}{4} \cdot \dfrac{1}{2} = \dfrac{W\sqrt{3}}{8}$

Finally, for the reaction force at peg $c$ ...

$\displaystyle \sum F_x = 0 \implies R_{cx} = T\sin{\theta} = \dfrac{W\sqrt{3}}{8}$

$\displaystyle \sum F_y = 0 \implies R_{cy} = T + T\cos{\theta} = T\left(1+\dfrac{\sqrt{3}}{2} \right) = \dfrac{W\sqrt{3}}{4}\left(\dfrac{2+\sqrt{3}}{2} \right) = W\left(\dfrac{2\sqrt{3}+3}{8}\right)$

$|R_c| = \sqrt{R_{cx}^2 + R_{cy}^2} = \sqrt{\dfrac{3W^2}{64} + \dfrac{W^2(21+12\sqrt{3})}{64}} = \dfrac{W\sqrt{6+3\sqrt{3}}}{4} \approx 0.84W$

That's about the best I can explain the method in this venue ... hope it clears up any confusion or misconceptions.