3. In this question i and i are perpendicular horizontal unit vectors.] Three forces. F_(1),F_(2) and F_(3) are given by F_(1)=(5i+2j)N F_(2)=(-3i+j)N F_(3)=(ai+bj)N where a and b are constants. The forces F_(1),F_(2) and F_(3) act on a particle P of mass 4kg. Given that P rests in equilibrium on a smooth horizontal surface under the action of these three forces. (a) find the size of the angle between the direction of F_(3) and the direction of -j (4) The force F_(3) is now removed and replaced by the force F_(4) given by F_(4)=lambda (i+3j)N where lambda is a positive constant. When the three forces F_(1),F_(2) and F_(4) act on P . the acceleration of P has magnitud 3.25ms^-2 (b) Find the value of lambda

(a) To find the size of the angle between the direction of $F_{3}$ and the direction of $-j$, we can use the dot product formula for vectors. The dot product of two vectors is given by:<br /><br />$F_{3} \cdot (-j) = (ai + bj) \cdot (-j) = -bj$<br /><br />The magnitude of $-j$ is 1, so the dot product simplifies to:<br /><br />$F_{3} \cdot (-j) = -b$<br /><br />The magnitude of $F_{3}$ is given by:<br /><br />$|F_{3}| = \sqrt{a^2 + b^2}$<br /><br />The cosine of the angle between $F_{3}$ and $-j$ is given by:<br /><br />$\cos(\theta) = \frac{F_{3} \cdot (-j)}{|F_{3}| \cdot |-j|} = \frac{-b}{\sqrt{a^2 + b^2}}$<br /><br />Therefore, the size of the angle between the direction of $F_{3}$ and the direction of $-j$ is:<br /><br />$\theta = \arccos\left(\frac{-b}{\sqrt{a^2 + b^2}}\right)$<br /><br />(b) When the three forces $F_{1}$, $F_{2}$, and $F_{4}$ act on P, the acceleration of P has magnitude $3.25ms^{-2}$. We can use Newton's second law of motion to find the value of $\lambda$.<br /><br />According to Newton's second law, the net force acting on an object is equal to the mass of the object multiplied by its acceleration. Mathematically, this can be expressed as:<br /><br />$F_{\text{net}} = m \cdot a$<br /><br />where $F_{\text{net}}$ is the net force, $m$ is the mass of the object, and $a$ is the acceleration.<br /><br />In this case, the net force is the sum of the three forces $F_{1}$, $F_{2}$, and $F_{4}$:<br /><br />$F_{\text{net}} = F_{1} + F_{2} + F_{4}$<br /><br />Substituting the given values, we have:<br /><br />$F_{\text{net}} = (5i + 2j)N + (-3i + j)N + \lambda (i + 3j)N$<br /><br />Combining like terms, we get:<br /><br />$F_{\text{net}} = (5 - 3 + \lambda)i + (2 + 1 + 3\lambda)j$<br /><br />According to Newton's second law, we have:<br /><br />$F_{\text{net}} = m \cdot a$<br /><br />where $m = 4kg$ and $a = 3.25ms^{-2}$.<br /><br />Substituting the given values, we have:<br /><br />$(5 - 3 + \lambda)i + (2 + 1 + 3\lambda)j = 4 \cdot 3.25ms^{-2}$<br /><br />Simplifying the equation, we get:<br /><br />$(\lambda + 2)i + (3\lambda + 3)j = 13ms^{-2}$<br /><br />Comparing the coefficients of $i$ and $j$ on both sides of the equation, we have:<br /><br />$\lambda + 2 = 13$<br /><br />$3\lambda + 3 = 0$<br /><br />Solving these equations, we find:<br /><br />$\lambda = 11$<br /><br />Therefore, the value of $\lambda$ is 11.