The children on the beam question(I'm assuming this is what you refer to as "painter question"):
Q3:
Sorry, but are the two lengths in figure3 1.0 and 3.5? I will assume so:
TO calculate the net torque I took downwards forces as negative, upwards as positive. My x axis had it's origin at the point where the pole touches the vertical beam and it goes from left to right.
The net torque is:
Where
is the Force acted by the wall.
Now this net torque simlifies to:
However the net torque must be zero if they are not rotating, hence:
30g-2.5F=0
30g=2.5F
F=12g N
The fact that F is positive indicates that the forces is upwards, since i chose up to be positive. So the force of the
wall on the beam is positive, however the
force of the beam on the wall is in the opposite direction(by Newton's third law) hence downwards, so B is the answer to q3 and u got the asnwer to 4
question 5:
Now in this case the you want the force the pole gives, that means x=0 cannot be at the pole, since 0*F=0.
Hence we will know make our origin at the wall:
Now the net torque is:
However the net torque must be zero. Hence:
and now solve for F.
question 6:
TO solve this question you have to look at how the x-value of the 40kg dude affects the net torque(which must be zero, hence
and
will vary)
Looking at our equation for
that was done in question 4:
We see that as x increases(he moves more towards the pole(hence to the right(which is positive))) the Net torque would decrease if F stayed constant. However we require the net torque to be zero so in order for that to happen F must decrease. So we know
decreases in magnitude.
Now to see what happens to
, look at our equation from the previous question:
In here, if x increases, the net torque would decrease if F stayed constant. However we want it to get back to 0, and in order for the to happen F must increase, so we know
increases, Hence the correct answer is C.
Edit: Fixed up the mistakes I made on question4 and 3 as stated in my latter post.