A crate of mass m is freely suspended at rest from a crane. When the operator begins to lift the crate further, the tension in the suspending cable increases uniformly from mg newtons to 1.2mg newtons over a period of 2 seconds. i) What is the tension in the cable t seconds after the lifting has

Question

A crate of mass m is freely suspended at rest from a crane. When the operator begins to lift the crate further, the tension in the suspending cable increases uniformly from mg newtons to 1.2mg newtons over a period of 2 seconds.
i) What is the tension in the cable t seconds after the lifting has begun (t ≤ 2)?
ii) What is the velocity after 2 seconds?
iii) How far has the crate risen after 2 seconds?
Assume the situation may be modelled with air resistance and cable stretching ignored.
Take g as 10 [latex]ms^{-2}[/latex].

Accepted Answer

When the crate is at rest it is in equilibrium and so the tension, T, in the cable equals the weight mg of the crate. After time t = 0, the tension increases, so there is a net upward force and the crate rises, see figure 18.1.

i) The tension increases uniformly by 0.2mg newtons in 2 seconds, i.e. it increases by 0.1mg newtons per second, see figure 18.2. After t seconds, the tension is

T = mg + 0.1mgt.

ii) As the force is a function of time use [latex]a = \frac{dv}{dt}[/latex]. Then at any moment in the 2-second period, F = ma gives

[latex]\begin{matrix} (mg + 0.1mgt) - mg = m\frac{dv}{dt} & \boxed{ ext{upwards is positive} }\end{matrix} [/latex]

⇒ [latex]\frac{dv}{dt} = 0.1gt.[/latex]

Integrating gives

v = 0.05gt² + k

where k is the constant of integration.
When t = 0, the crate has not quite begun to move, so v = 0. This gives k = 0 and v = 0.05gt².

When t is 2,

v = 0.5 × 10 × 4

= 2.

The velocity after 2 seconds is 2 [latex]ms^{-1}[/latex].

iii) To find the displacement s, write v as [latex]\frac{ds}{dt}[/latex] and integrate again.

[latex]\frac{ds}{dt} = 0.05gt²[/latex]

s = 0.05gt²dt

s = 0.05g × [latex]\frac{1}{3}t³[/latex] + c

When t = 0, s = 0 ⇒ c = 0.

When t = 2 and g = 10, s = [latex]\frac{4}{3}.[/latex]

The crate moves [latex]\frac{4}{3}[/latex] m in 2 seconds.

Question 18.1: A crate of mass m is freely suspended at rest from a crane. ...

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