Question 296: Centre of Mass and Rotational Motion

Question

A slab is subjected to two forces $$\overrightarrow {{F_1}} $$ and $$\overrightarrow {{F_2}} $$ of  same magnitude F as shown in the figure. Force $$\overrightarrow {{F_2}} $$ is in XY-plane while force $$\overrightarrow {{F_1}} $$ acts along z = axis at the point $$\left( {2\overrightarrow i  + 3\overrightarrow j } \right).$$. The moment of these forces about point O will be :

Accepted Answer

To determine the moment of the forces about point O, we need to calculate the moments of each force and then combine them vectorially.
First, consider force $$\overrightarrow{F_1}$$. This force acts along the z-axis at the point $$(2\overrightarrow{i} + 3\overrightarrow{j})$$.
The position vector of the point of application of $$\overrightarrow{F_1}$$ relative to the origin O is:
$$\overrightarrow{r_1} = 2\overrightarrow{i} + 3\overrightarrow{j}$$
Since $$\overrightarrow{F_1}$$ acts along the z-axis, we can express it as:
$$\overrightarrow{F_1} = F\overrightarrow{k}$$
The moment of force $$\overrightarrow{F_1}$$ about point O is given by the cross product:
$$\overrightarrow{M_1} = \overrightarrow{r_1} 	imes \overrightarrow{F_1}$$
Substitute the values:
$$\overrightarrow{M_1} = (2\overrightarrow{i} + 3\overrightarrow{j}) 	imes F\overrightarrow{k}$$
Compute the cross product:
$$\overrightarrow{M_1} = 2F(\overrightarrow{i} 	imes \overrightarrow{k}) + 3F(\overrightarrow{j} 	imes \overrightarrow{k})$$

$$\overrightarrow{M_1} = 2F(-\overrightarrow{j}) + 3F\overrightarrow{i}$$

$$\overrightarrow{M_1} = -2F\overrightarrow{j} + 3F\overrightarrow{i}$$
So, the moment due to force $$\overrightarrow{F_1}$$ is:
$$\overrightarrow{M_1} = 3F\overrightarrow{i} - 2F\overrightarrow{j}$$
Torque for F
2
force
$${\overrightarrow F _2}$$ = $${F \over 2}\left( { - \widehat i} ight) + {{F\sqrt 3 } \over 2}\left( { - \widehat j} ight)$$
$${\overrightarrow r _1} = 0\widehat i + 6\widehat j$$
$${\overrightarrow 	au  _{_{{F_2}}}} = \overrightarrow {{r_1}}  	imes \overrightarrow {{F_1}}  = 3F\widehat k$$
$${\overrightarrow 	au  _{net}} = {\overrightarrow 	au  _{{F_1}}} + {\overrightarrow 	au  _{{f_2}}}$$
$$ = 3F\widehat i + 2F\left( { - \widehat j} ight) + 3F\left( {\widehat k} ight)$$

Answer

$$\left( {3\widehat i - 2\widehat j - 3\widehat k} \right)F$$

Answer

$$\left( {3\widehat i + 2\widehat j - 3\widehat k} \right)F$$

Answer

$$\left( {3\widehat i + 2\widehat j + 3\widehat k} \right)F$$

Answer

$$\left( {3\widehat i - 2\widehat j + 3\widehat k} \right)F$$

Centre of Mass and Rotational Motion

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