required information: the two couples shown are to be replaced with a single equivalent couple. given: f1 =…

Explanation:

Step1: Calculate moment of couple 1

The moment of a couple $\vec{M}$ is given by $\vec{M}=\vec{r}\times\vec{F}$. For the couple with force $F_1 = 29N$, the perpendicular - distance vector $\vec{r}_1$ between the two forces of the couple. The moment of the couple $M_1$ due to $F_1$ is $M_1 = F_1d_1$, where $d_1=0.75m$. So, $M_1 = 29\times0.75\sin40^{\circ}$. [M_1=29\times0.75\times0.6428 = 29\times0.4821=13.9809N\cdot m]

Step2: Calculate moment of couple 2

For the couple with force $F_2 = 24N$, the perpendicular - distance vector $\vec{r}_2$ between the two forces of the couple. The moment of the couple $M_2$ due to $F_2$ is $M_2 = F_2d_2$, where $d_2 = 0.3m$. So, $M_2=24\times0.3$. [M_2 = 7.2N\cdot m]

Step3: Find the resultant couple moment

Since the two couples act in the same plane (assumed from the problem setup), the resultant couple moment $M_R$ is the sum of the two couple moments. $M_R = M_1+M_2$. [M_R=13.9809 + 7.2=21.1809N\cdot m\approx21.18N\cdot m]

Answer:

$21.18N\cdot m$