Finding a component of a force given the resultant magnitude

PhysicsVector MechanicsMediumJEE

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1. If the resultant of three forces $\vec{F}_1 = p\hat{i} + 3\hat{j} - \hat{k}$, $\vec{F}_2 = -5\hat{i} + \hat{j} + 2\hat{k}$ and $\vec{F}_3 = 6\hat{i} - \hat{k}$ acting on a particle has magnitude equal to 5 units, then the value (s) of p is (are) (A) -6 (B) -4 (C) 2 (D) 4

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Step 1

Hi Gnyanesh, let's solve this physics problem together. We are given three force vectors acting on a particle, and we know that the magnitude of their resultant is equal to five units. We need to find the possible values of the parameter p.

Given Force Vectors

$$\vec{F}_1 = p\hat{i} + 3\hat{j} - \hat{k}$$
$$\vec{F}_2 = -5\hat{i} + \hat{j} + 2\hat{k}$$
$$\vec{F}_3 = 6\hat{i} - \hat{k}$$
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Step 2

The resultant force vector, which we will call F net, is the vector sum of these three individual forces.

$$\vec{F}_{\text{net}} = \vec{F}_1 + \vec{F}_2 + \vec{F}_3$$
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Step 3

Let us substitute the given expressions for each force vector into this equation.

$$\vec{F}_{\text{net}} = (p\hat{i} + 3\hat{j} - \hat{k}) + (-5\hat{i} + \hat{j} + 2\hat{k}) + (6\hat{i} - \hat{k})$$
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Step 4

Now, let's group the components along the i, j, and k unit vectors.

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Step 5

Simplifying the coefficients, we get p plus one along the i direction, four along the j direction, and zero along the k direction.

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Step 6

Now, let's find the magnitude of this resultant force. The magnitude of a vector is the square root of the sum of the squares of its components.

Magnitude of the Resultant Force

$$|\vec{F}_{\text{net}}| = \sqrt{F_x^2 + F_y^2 + F_z^2}$$
$$|\vec{F}_{\text{net}}| = \sqrt{(p + 1)^2 + 4^2 + 0^2}$$
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Step 7

We are given that this magnitude is equal to five units, so we can equate this expression to five.

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About This Question

Subject
Physics
Topic
Vector Mechanics
Difficulty
Medium
Exam
JEE
Question Type
Multiple Choice

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