Question

A force of 500 N is applied to a wrench as shown in the figure. The force is being applied at an angle of ${235}^{\circ }$ with positive x – axis

The length of the wrench is 0.3 m The magnitude of vector moment of the force F acting at A about origin is given by _____ and the screw at origin will move in ____ direction.

• A. $500\times 0.3\times sin{235}^{\circ }$ negative z direction
• B. $500\times 0.3\times sin{55}^{\circ }$ positive z direction
• C. $500\times 0.3\times sin{125}^{\circ }$ negative z direction
• D. $500\times 0.3\times sin{135}^{\circ }$positive z direction

Answer 1

Answer: (B), (D)

$500\times 0.3\times sin{135}^{\circ }$positive z direction

We know that vector moment of a force F acting at a point is given by $\overrightarrow{r}\times \overrightarrow{F}$ Here.$\overrightarrow{r}$ is the position vector of the point of application of the force with respect to origin. In the given figure$\overrightarrow{r}$ will point from origin towards point $\overrightarrow{A}$ as shown in the following diagram

Now the angle between $\overrightarrow{r}$ and $\overrightarrow{F}$ will be $\varphi$ which is nothing but ${235}^{\circ }-{180}^{\circ }={55}^{\circ }$
So magnitude of $\overrightarrow{r}\times \overrightarrow{F}$ will be $|\overrightarrow{r}||\overrightarrow{F}|sin\varphi$
=$0.3\times 500\times sin\left({55}^{\circ }\right)$ (Here $|\overrightarrow{r}|$ = 0.3 m because the length of the wrench in 0.3 m and the force is being applied at the end of the wrench. So point of application of force is 0.3 m from the origin, which is nothing but the modulus of the position vector of that point)
=$0.3\times 500\times sin\left({125}^{\circ }\right)$ (Because sin $\left(\varphi \right)$ =sin$(180-\varphi )$
For the direction of $\overrightarrow{r}\times \overrightarrow{F}$ which is same as the direction in which screw at origin will move, will be given by right hand rule, which comes out to be the z direction in this case.