Physics 9702 · AS & A Level · Turning effects of forces

Turning effects of forces — practice question

(a)[1]

State the meaning of the centre of gravity of an object.

(b)

As shown in Fig. 2.1, a non-uniform rod XY is pivoted at point P. The rod is $4.00\,\text{m}$ long and has weight $44.0\,\text{N}$. Its centre of gravity is $1.70\,\text{m}$ from end X of the rod. Point P is $1.10\,\text{m}$ from end X. A sphere is suspended from end Y by a wire. The sphere weighs $3.0\,\text{N}$. The wire’s weight is negligible. A vertical downward force $F$ acts at end X so that the horizontal rod is in equilibrium.

(b(i))[3]

Calculate $F$ by taking moments about P.

(b(ii))[1]

Calculate the force that the pivot exerts on the rod.

(c)

In (b), the sphere is now placed in a liquid in a container, as shown in Fig. 2.2. The density of the liquid is $1100\,\text{kg m}^{-3}$. The upthrust acting on the sphere due to the liquid is $2.5\,\text{N}$. The size of $F$ is unchanged, so the horizontal rod is not in equilibrium.

(c(i))[3]

Use Archimedes’ principle to work out the radius $r$ of the sphere.

(c(ii))[2]

Calculate the magnitude and direction of the resultant moment of the forces on the rod about P.

Worked solution & mark scheme

This 10-mark question has a full step-by-step worked solution and mark scheme. One marking point: the point at which the object’s entire weight is considered to act

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