Solution

Moment of $F$ about $P$ is $F \times d sin \theta$ clockwise. Here, we have found the perpendicular distance from $P$ using trigonometry and multiplied it by the magnitude of the force. When describing a moment you need to give a direction of rotation. Therefore when we are given the values this is $$\begin{align} \text{Moment} & = F \times d \sin \theta, \\ & = 8 \times 14\sin(35^{\circ}), \\ &= 64.241\mathrm{Nm} \text{ (to 3d.p.).} \end{align}$$ The moment of the force is measured in newton-metres $\mathrm{N m}$, therefore the moment of $F$ about $P$ is $64.241\mathrm{Nm}$.

Solution

Each force is already perpendicular to the point $P$.

The moment of the $6\mathrm{N}$ force is $6 \times 2 = 12 \mathrm{N m}$ anticlockwise.

The moment of the $14\mathrm{N}$ force is $14 \times 2 = 28 \mathrm{N m}$ clockwise.

The moment of the $5\mathrm{N}$ force is $5 \times (2+3) = 25 \mathrm{Nm}$ anticlockwise.

Total clockwise $= 28 \mathrm{N m}$ and total anticlockwise $= 37 \mathrm{Nm}$. Therefore the sum of the moments is $37 - 28 = 9 \mathrm{Nm}$ anticlockwise. As the anticlockwise total was greater we chose anticlockwise as the positive direction.

Solution

The perpendicular distance from $P$ to the force will be the distance opposite the angle given by $7 \sin (60^{\circ})$. Therefore $$\begin{align} \text{The moment of the force} & = 11 \times 7 \sin (60^{\circ}), \\ & = 77 \sin (60^{\circ}), \\ & = 66.684 \mathrm{Nm} \text{ anticlockwise (3 d.p.).} \end{align}$$

Solution

We can draw a diagram showing all the forces acting.

The weight of the rod will act at its centre of mass - as it is a uniform rod the weight acts at the mid-point. The rod is in equilibrium so the total forces acting upwards will equal the total forces acting downwards when we resolve vertically. $$\begin{equation} R_X + R_Y = 30. \end{equation}$$ Consider the moments about point $X$, we have that clockwise moments will equal anticlockwise moments therefore $$\begin{align} 30 \times 4.5 & = R_Y \times (4.5 + 0.5), \\ 135 & = 5R_Y, \\ 27 \mathrm{N} & = R_Y. \end{align}$$ Now we can use this value to find $R_X$ $$\begin{align} R_X + R_Y & = 30, \\ R _ X & = 30 - 27, \\ & = 3 \mathrm{N}. \end{align}$$ Therefore the reaction at point $X$ is $3 \mathrm{N}$ and the reaction at point $Y$ is $27\mathrm{N}$.

Solution

We can draw a diagram showing all the forces acting.

Here we have assumed that the centre of mass acts at a point $x$ m from point $A$. As the rod is not uniform we can not say that the centre of mass is at the mid point of the rod. We resolve vertically $$\begin{align} 4R + R & = 15, \\ 5R & = 15, \\ R & = 3. \end{align}$$ Now we take moments about point $A$ $$\begin{align} 15 \times x & = (4R \times 3) + (R \times 7), \\ & = 36 + 21, \\ & = 57, \\ x & = 3.8. \end{align}$$ The centre of mass is $3.8 \mathrm{m}$ from $A$.