# Thread: algebra in mechanics

1. ## algebra in mechanics

im not sure where to post this cos it includes mechanics formulae but im sure most of you have done mechanics as well since it is often included in maths programmes
so

a load of weight 7 kN is being raised from rest with constant acceleration by a cable. after the load has been raised 20 metres, the cable suddenly becomes slack. the load continues upwards for a distance of 4 metres before coming to instantaneous rest. Assuming no air resistance, find the tension in the cable before it became slack.

so obviously T = 7000 N + a(700), a being acceleration
but i can't seem to find this acceleration; ive tried using a velocity-time graph but there are too many unknowns for me; i am also a bit confused by what is meant by 'instantaneous' rest: does this mean constant velocity for 4 metres and then boom the graph line goes vertical?

2. When the load is raised to 20 m it acquires a velocity v. When the cable becomes slack, the load continues its upward motion with a retardation g. Its instantaneous final velocity is zero. During that time it moves 4 m up. Using relevant kinematic equation find the velocity v.
This velocity v is the final velocity for the first part of the problem.
In that initial velocity is zero. Final velocity we found in the second part. Displacement is given. Find the acceleration. From that you can find the tension in the cable.

3. so the load the 'instantaneous' rest is caused by g? this is the constant deceleration over the four metres? i.e the velocity-time sketch of this portion of the movement will be a triangle and not a trapezium, am i correct?

4. Hello furor celtica
Originally Posted by furor celtica
im not sure where to post this cos it includes mechanics formulae but im sure most of you have done mechanics as well since it is often included in maths programmes
so

a load of weight 7 kN is being raised from rest with constant acceleration by a cable. after the load has been raised 20 metres, the cable suddenly becomes slack. the load continues upwards for a distance of 4 metres before coming to instantaneous rest. Assuming no air resistance, find the tension in the cable before it became slack.

so obviously T = 7000 N + a(700), a being acceleration
but i can't seem to find this acceleration; ive tried using a velocity-time graph but there are too many unknowns for me; i am also a bit confused by what is meant by 'instantaneous' rest: does this mean constant velocity for 4 metres and then boom the graph line goes vertical?
I assume from the 700 in your equation that you've been told to take $\displaystyle g = 10$. I've done the same below. If you didn't mean this, and you should have taken $\displaystyle g = 9.8$ (which is more usual), then you'll have to make the necessary amendments.

For both parts of the motion, we need to connect final and initial velocities, acceleration and distance. So, with the usual notation, the equation we need is:
$\displaystyle v^2 = u^2 + 2as$
In the second part of the motion, we have:
$\displaystyle v = 0$ (the body comes to rest)

$\displaystyle a = -g = -10$ (see above)

$\displaystyle s=4$
So, if $\displaystyle V$ is the velocity at the start of phase 2:
$\displaystyle 0=V^2+2(-10)(4)$

$\displaystyle \Rightarrow V^2 = 80$
But, of course, $\displaystyle V$ is also the velocity at the end of phase 1. During this phase, then:
$\displaystyle u = 0$

$\displaystyle v = V$

$\displaystyle s = 20$
and we need to find $\displaystyle a$. So:
$\displaystyle V^2 = 0^2 +2a(20)$

$\displaystyle \Rightarrow 40a = 80$

$\displaystyle \Rightarrow a = 2$
Now we can use the equation of motion:
$\displaystyle \sum F = ma$
with $\displaystyle m = 700$ and $\displaystyle a = 2$
$\displaystyle \Rightarrow T - 7000 = 700\cdot2$

$\displaystyle \Rightarrow T = 8400$
So the tension in the string is $\displaystyle 8400$ N.