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A square $ABCD$ is actually a central cross-section of a cuboidal block of weight $W$. The block stands on the ground with $AD$ against a vertical wall. The edge through $A$ perpendicular to $ABCD$ is smoothly hinged to the wall. A cable is attached to the block at $B$, and this is used to raise the block slowly off the ground until $AB$ is vertical. The cable passes over a pulley at the point $E$ above $A$ so that $AE = AB$, and the other end is wound onto a drum powered by a motor. Show that, when $AB$ makes an angle $θ$ with the horizontal, the tension in the cable is

$$\frac{1}{2}√2W\frac{sin(45^o+θ)}{sin(45^o+\frac{1}{2}θ)}$$

Suggested answer:

Let the length of a side of the cube be $l$

Taking moments about the hinge (which is all along the edge of the cube attached to the wall):

$$Wlsin(45^o+θ) = Tlsin(45^o+θ)+Tlcos(45^o+θ)$$ which gives, using trig identities ($sin(A+B)=sinAcosB+cosAsinB$ and $cos(A+B)=cosAcosB-sinAsinB)$ for $sin(45^o+θ)$ and $cos(45^o+θ)$

$$Wlsin(45^o+θ) = Tl\frac{2}{√2}cosθ$$ which gives

$$Wlsin(45^o+θ) = Tl\frac{2}{√2}sin(90^o-θ)$$ or

$$Wlsin(45^o+θ) = Tl\frac{2}{√2}sin(45^o+\frac{1}{2}θ)$$

So, cancelling $l$ and rearranging gives

$$\frac{1}{2}√2W\frac{sin(45^o+θ)}{sin(45^o+\frac{1}{2}θ)}$$

But can you say that the weight $W$ is a distance $l$ from the hinge rather than $\frac{l}{2}$, which would be at the centre of mass of the block?

The problem is that of you take moments about the hinge with the weight at $\frac{l}{2}$, you end up with

$$\frac{1}{4}√2W\frac{sin(45^o+θ)}{sin(45^o+\frac{1}{2}θ)}$$

J132
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1 Answers1

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There are two torques acting on the system, one from the cable and the other from gravity .

Taking $A$ as the pivot we have the cable tension $T$ acting at an angle of $45^o + \frac\theta 2$ on the arm $AB$ which has length $l$

$$\tau_{cable}=Tl\sin(45^o + \frac\theta 2)$$

The balancing torque comes from the weight of the block which acts as if it is concentrated at the midpoint of $AC$ ,a distance $\frac{l}{\sqrt 2}$ from $A$. It is best to draw diagram to work out the angles, but you should get that $AC$makes an angle $\alpha = 135^o - \theta$ with the vertical.

$$\tau_{gravity}=\frac{Wl}{\sqrt 2}\sin(135^o -\theta )=\frac{Wl}{\sqrt 2}\sin(45^o +\theta )$$

The required expression for $T$ is obtained by setting $\tau_{gravity}=\tau_{cable}$

WW1
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  • Thanks for that. Yes, I was looking at the diagonal, $AC$, through the cube as well at one point, when taking moments about the hinge, as when the block is moved up half way (when angle $θ= 45^o$), $AC$ will be horizontal with length $√2l$. Also, the angle $AC$ makes to the vertical is $180^o - 45^o = 135^o$ when $θ= 0^o$. Then when $θ>0^o$, the angle that $AC$ makes to the vertical is $135^o-θ$, as the block is raised. – J132 Oct 07 '15 at 22:32