Question

A string is pulled with a force F = 130 N as shown. The force vector comes out to be

$\overrightarrow{F}$ = 10 ($\alpha \hat{i}$ + $\beta \hat{j}$ + $\gamma \hat{k}$) N. Write $\beta$ + $\alpha$ + $\gamma$ in OMR sheet.

Answer 1

13

Answer 2

The force vector $\vec{F}$ is directed from the origin O(0, 0, 0) to the point P(12, 4, -3).

The position vector of point P relative to the origin is $\vec{r} = \vec{OP} = (12-0)\hat{i} + (4-0)\hat{j} + (-3-0)\hat{k} = 12\hat{i} + 4\hat{j} - 3\hat{k}$.

The magnitude of this position vector is $|\vec{r}| = \sqrt{12^2 + 4^2 + (-3)^2} = \sqrt{144 + 16 + 9} = \sqrt{169} = 13$.

The unit vector in the direction of $\vec{F}$ is $\hat{r} = \frac{\vec{r}}{|\vec{r}|} = \frac{12\hat{i} + 4\hat{j} - 3\hat{k}}{13}$.

The magnitude of the force is given as $F = 130$ N.

The force vector $\vec{F}$ can be written as the product of its magnitude and the unit vector in its direction: $\vec{F} = F \hat{r} = 130 \times \frac{12\hat{i} + 4\hat{j} - 3\hat{k}}{13}$. $\vec{F} = 10 \times (12\hat{i} + 4\hat{j} - 3\hat{k}) = 120\hat{i} + 40\hat{j} - 30\hat{k}$.

The problem states that the force vector is given by $\vec{F}$ = 10 ($\alpha \hat{i}$ + $\beta \hat{j}$ + $\gamma \hat{k}$) N.

Comparing this form with the calculated force vector: $10 (\alpha \hat{i} + \beta \hat{j} + \gamma \hat{k}) = 10 (12\hat{i} + 4\hat{j} - 3\hat{k})$.

Dividing by 10 on both sides: $\alpha \hat{i}$ + $\beta \hat{j}$ + $\gamma \hat{k}$ = $12\hat{i} + 4\hat{j} - 3\hat{k}$.

Comparing the coefficients of $\hat{i}$, $\hat{j}$, and $\hat{k}$: $\alpha = 12$ $\beta = 4$ $\gamma = -3$

We need to find the value of $\beta + \alpha + \gamma$. $\beta + \alpha + \gamma = 4 + 12 + (-3) = 16 - 3 = 13$.