So applying Newton's second law I ended up with:
(Fth / m) - g = a
But m is dependent on time so that becomes:
[Fth / (m0 - R * t)] - g = a
Do I then have to integrate that somehow?
Take a look at this:
The angle 52* West of North results in line c (note that in my pretty drawing, angle C and side c have no relation). Angle C is 52*, and since the axis are perpindicular, angle A + angle C needs to equal 90*, so A is equal to 38*.
So then cos(A) = b / c -> b = c *...
Homework Statement
Prove that the upward velocity of a rocket of initial mass M0, which is propelled by fuel burning at a rate of R kg/s, is given by vy = uex * ln[M0 / M(t) ] - g * t. uex is the speed of the exhaust gas relative to the rocket and M0 is the initial mass (rocket + fuel).
Also...
The angle is 52 degrees west of north. So you start at north and rotate 52 degrees to the left. The triangle will look similar to this:
Angle B -> |\
| \
|__\ <- Angle A
Angle A is 38* and Angle B is 52*.
Hope this helps.
That is what F equals though, isn't it? The question asks the max force that can be applied to the block so it doesn't slide. The max force that the static friction can apply is 39.24N, and so the max force on the string that can be applied is also 39.24N.
Homework Statement
A 10kg block is resting on a 5kg bracket, which rests on a frictionless surface. The coef. of static and kinetic friction between the block and bracket are .4 and .3, respectively. Find out a) the max force that can be applied to the block without the block sliding on the...