Law of Parallelogram of Forces : 5 in 5 MCQs S01-E01

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As solved in the video under 5 minutes, this is the notes for the 5 MCQs solved with explanation. MCQ type questions are not only helpful for your examinations but also for competitive exams like GATE, UPSC and many more.

What is Law of Parallelogram of Forces?

law-of-parallelogram-of-forces-diagram

Consider two forces PP and QQ acting simultaneously on a particle OO as shown in the figure above. Let the angle between these two forces be θ\theta. By extending imaginary lines from PP and QQ, we can get a parallelogram as shown in the figure above as dotted lines.

Therefore, by the Law of Parallelogram of Forces the resultant of the forces PP and QQ can be represented by the diagonal of this parallelogram and its direction as α°\alpha \degree from force PP.

Definition of Law of Parallelogram of Forces

If two forces acting simultaneously on a particle are represented in magnitude and direction by the two sides of a parallelogram, then their resultant can be represented in magnitude and direction by the diagonal of the parallelogram which passes through their point of intersection.

And the magnitude of the resultant is given by:

R=P2+Q2+2 PQ cosθR = \sqrt{P^2 + Q^2 + 2 \ PQ \ \cos \theta}

And the direction of the resultant is given by:

tanα=Q sinθP+Q cosθ\tan \alpha = {Q \ \sin \theta \over P + Q \ \cos \theta}

Now, let's start solving MCQs. These questions will have 2 forces acting on a particle, hence we will apply the law of parallelogram of forces.

MCQ - 1

If P>QP \gt Q, and PP and QQ are acting along a same straight line, but in opposite direction. Then the resultant is?

a) P+QP+Q              c) PQP \over Q

b) PQP-Q              d) QPQ \over P

Given:
θ=180°\theta = 180 \degree (because P and Q are in straight line, but opposite in direction)

R=P2+Q2+2 PQ cosθ=P2+Q2+2 PQ cos180=P2+Q22 PQ=(PQ)2R=PQ\begin{aligned} R &= \sqrt{P^2 + Q^2 + 2 \ PQ \ \cos \theta} \\ &= \sqrt{P^2 + Q^2 + 2 \ PQ \ \cos 180} \\ &= \sqrt{P^2 + Q^2 - 2 \ PQ} \\ &= \sqrt{(P - Q)^2} \\ & \boxed{\therefore R = P - Q} \\ \end{aligned}

Answer: b


MCQ - 2

The resultant of two equal forces P making an angle of θ\theta is:

a) 2Psinθ22P \sin {\theta \over 2}              c) 2Ptanθ22P \tan {\theta \over 2}

b) 2Pcosθ22P \cos {\theta \over 2}              d) 2Pcotθ22P \cot {\theta \over 2}

Given:
P=QP = Q

R=P2+Q2+2 PQ cosθ=P2+P2+2 PP cosθ=2P2+2 P2cosθ=2P2(1+cosθ)=2P2(2cos2θ2)R=2Pcosθ2\begin{aligned} R &= \sqrt{P^2 + Q^2 + 2 \ PQ \ \cos \theta} \\ &= \sqrt{P^2 + P^2 + 2 \ P \sdot P \ \cos \theta} \\ &= \sqrt{2P^2 + 2 \ P^2 \cos \theta} \\ &= \sqrt{2P^2 (1 + \cos \theta)} \\ &= \sqrt{2P^2 (2 \cos^2 {\theta \over 2})} \\ &\boxed{\therefore R = 2P \cos {\theta \over 2}} \end{aligned}

Answer: b




MCQ - 3

The angle between two forces when resultant is maximum and minimum respectively is:

a) 0°&180°0 \degree \And 180 \degree              c) 90°&180°90 \degree \And 180 \degree

b) 180°&0°180 \degree \And 0 \degree              d) 90°&0°90 \degree \And 0 \degree

Explanation:
In the question it is asked to find the value of θ\theta for the maximum value RR and the minimum value of RR.

We know that: R=P2+Q2+2 PQ cosθR = \sqrt{P^2 + Q^2 + 2 \ PQ \ \cos \theta}.
Now if we change the value of cosθcos \theta and add or subtract it from the remaining value of PP and QQ we get the value of RR.

So, to find the answer, let's take 4 extreme values from the unit circle as given in the options for θ0°, 90°, 180° and 270°\theta \to 0 \degree, \ 90 \degree, \ 180 \degree \ and \ 270 \degree
Now the cos\cos values for these angles will be 1,0,1 and 0\to 1, 0, -1 \ and \ 0 respectively.

Here, maximum value is 11(i.e. cos0\cos 0) and minimum value is 1-1(i.e. cos180\cos 180).
Therefore, at 0°0 \degree and at 180°180 \degree, the resultant will be maximum and minimum respectively.

Answer: a




MCQ - 4

The resultant of two forces each equal to P and acting at right angle is:

a) P2{P \over 2}              c) P22{P \over 2 \sqrt2}

b) P2{P \over \sqrt2}              d) 2P\sqrt{2} \sdot P

Given:
P=QP = Q, θ=90°\theta = 90\degree

R=P2+Q2+2 PQ cosθ=P2+Q2+2 PQ cos90=P2+P2=2P2R=2P\begin{aligned} R &= \sqrt{P^2 + Q^2 + 2 \ PQ \ \cos \theta} \\ &= \sqrt{P^2 + Q^2 + 2 \ PQ \ \cos 90} \\ &= \sqrt{P^2 + P^2} \\ &= \sqrt{2P^2} \\ &\boxed{R = \sqrt{2} \sdot P} \\ \end{aligned}

Answer: d




MCQ - 5

The resultant of two forces PP and QQ is RR. If QQ is doubled, the new resultant is perpendicular to P. Then,

a) P=Q{P = Q}              c) Q=R{Q = R}

b) Q=2R{Q = 2R}              d) none of these\text{none of these}

Explanation:
We have two set of data here:

  1. Forces PP and QQ, whose resultant is R
  2. If QQ is doubled, then resultant(R') would be perpendicular to PP. This means α=90°\alpha = 90 \degree and Q=2QQ = 2Q
For α=90°, Q=2Q we know thattanα=Q sinθP+Q cosθtan90=2QsinθP+Q cosθ=2QsinθP+Q cosθ\begin{aligned} For \ \alpha &= 90 \degree, \ Q = 2Q \ \text{we know that} \\ \\ \tan \alpha &= {Q \ \sin \theta \over P + Q \ \cos \theta} \\ \\ \therefore \tan 90 &= {2Q \sin \theta \over P + Q \ \cos \theta} \\ \\ \therefore \infin &= {2Q \sin \theta \over P + Q \ \cos \theta} \\ \end{aligned}

This is only possible if:

P+Q cosθ=0P=2Qcosθ(1)\begin{aligned} &P + Q \ \cos \theta = 0 \\ &\boxed{\therefore P = -2Q \cos \theta} ---- (1) \end{aligned}

Using the formula for resultant force:

R=P2+Q2+2 PQ cosθUsing from equation (1)=(2Qcosθ)2+Q2+2(2Qcosθ)Qcosθ=4Q2cos2θ+Q24Q2cos2θ=Q2R=Q\begin{aligned} R &= \sqrt{P^2 + Q^2 + 2 \ PQ \ \cos \theta} \\ &\text{Using from equation (1)} \\ &= \sqrt{(-2Q \cos \theta)^2 + Q^2 + 2(-2Q \cos \theta) \sdot Q cos \theta} \\ &= \sqrt{4Q^2 \cos^2 \theta + Q^2 - 4Q^2 \cos^2 \theta} \\ &= \sqrt{Q^2} \\ &\boxed{\therefore R = Q} \end{aligned}

Answer: c

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