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Question: Two forces P and Q of magnitude 2F and 3F, respectively, are at an angle $\theta$ with each other. I...

Two forces P and Q of magnitude 2F and 3F, respectively, are at an angle θ\theta with each other. If the forces are doubled, then their resultant also gets doubled. Then, the angle is :

A

30°

B

60°

C

90°

D

120°

Answer

120°

Explanation

Solution

The problem involves finding the angle between two forces given initial conditions and how the resultant changes when the forces are modified.

Let the initial forces be P1=2FP_1 = 2F and P2=3FP_2 = 3F. Let the angle between them be θ\theta. The magnitude of their resultant, RR, is given by the formula: R2=P12+P22+2P1P2cos⁡θR^2 = P_1^2 + P_2^2 + 2P_1 P_2 \cos \theta

Substituting the given magnitudes: R2=(2F)2+(3F)2+2(2F)(3F)cos⁡θR^2 = (2F)^2 + (3F)^2 + 2(2F)(3F) \cos \theta R2=4F2+9F2+12F2cos⁡θR^2 = 4F^2 + 9F^2 + 12F^2 \cos \theta R2=13F2+12F2cos⁡θR^2 = 13F^2 + 12F^2 \cos \theta --- (Equation 1)

The problem states that "If the forces are doubled, then their resultant also gets doubled".

If both forces were doubled (P1→4FP_1 \to 4F and P2→6FP_2 \to 6F), the new resultant R′R' would be: (R′)2=(4F)2+(6F)2+2(4F)(6F)cos⁡θ(R')^2 = (4F)^2 + (6F)^2 + 2(4F)(6F) \cos \theta (R′)2=16F2+36F2+48F2cos⁡θ(R')^2 = 16F^2 + 36F^2 + 48F^2 \cos \theta (R′)2=52F2+48F2cos⁡θ(R')^2 = 52F^2 + 48F^2 \cos \theta Factoring out 4: (R′)2=4(13F2+12F2cos⁡θ)(R')^2 = 4(13F^2 + 12F^2 \cos \theta)

From Equation 1, we know 13F2+12F2cos⁡θ=R213F^2 + 12F^2 \cos \theta = R^2. So, (R′)2=4R2(R')^2 = 4R^2, which means R′=2RR' = 2R.

This shows that if both forces are scaled by a factor, the resultant is also scaled by the same factor. This condition is always true, regardless of the angle θ\theta. Therefore, the problem cannot be solved if "the forces are doubled" means both forces are doubled.

In such ambiguous questions, it is common to interpret "the forces are doubled" as "one of the forces is doubled". We test both possibilities:

Case 1: Force P (2F) is doubled to 4F, and Force Q (3F) remains unchanged. New forces are P1′=4FP_1' = 4F and P2′=3FP_2' = 3F. New resultant R′=2RR' = 2R. (R′)2=(P1′)2+(P2′)2+2P1′P2′cos⁡θ(R')^2 = (P_1')^2 + (P_2')^2 + 2P_1' P_2' \cos \theta (2R)2=(4F)2+(3F)2+2(4F)(3F)cos⁡θ(2R)^2 = (4F)^2 + (3F)^2 + 2(4F)(3F) \cos \theta 4R2=16F2+9F2+24F2cos⁡θ4R^2 = 16F^2 + 9F^2 + 24F^2 \cos \theta 4R2=25F2+24F2cos⁡θ4R^2 = 25F^2 + 24F^2 \cos \theta --- (Equation 2a)

Substitute R2R^2 from Equation 1 into Equation 2a: 4(13F2+12F2cos⁡θ)=25F2+24F2cos⁡θ4(13F^2 + 12F^2 \cos \theta) = 25F^2 + 24F^2 \cos \theta 52F2+48F2cos⁡θ=25F2+24F2cos⁡θ52F^2 + 48F^2 \cos \theta = 25F^2 + 24F^2 \cos \theta 48F2cos⁡θ−24F2cos⁡θ=25F2−52F248F^2 \cos \theta - 24F^2 \cos \theta = 25F^2 - 52F^2 24F2cos⁡θ=−27F224F^2 \cos \theta = -27F^2 cos⁡θ=−2724=−98\cos \theta = -\frac{27}{24} = -\frac{9}{8}

This value is less than -1, which is physically impossible for cos⁡θ\cos \theta. So, this case is not the intended one.

Case 2: Force Q (3F) is doubled to 6F, and Force P (2F) remains unchanged. New forces are P1′=2FP_1' = 2F and P2′=6FP_2' = 6F. New resultant R′=2RR' = 2R. (R′)2=(P1′)2+(P2′)2+2P1′P2′cos⁡θ(R')^2 = (P_1')^2 + (P_2')^2 + 2P_1' P_2' \cos \theta (2R)2=(2F)2+(6F)2+2(2F)(6F)cos⁡θ(2R)^2 = (2F)^2 + (6F)^2 + 2(2F)(6F) \cos \theta 4R2=4F2+36F2+24F2cos⁡θ4R^2 = 4F^2 + 36F^2 + 24F^2 \cos \theta 4R2=40F2+24F2cos⁡θ4R^2 = 40F^2 + 24F^2 \cos \theta --- (Equation 2b)

Substitute R2R^2 from Equation 1 into Equation 2b: 4(13F2+12F2cos⁡θ)=40F2+24F2cos⁡θ4(13F^2 + 12F^2 \cos \theta) = 40F^2 + 24F^2 \cos \theta 52F2+48F2cos⁡θ=40F2+24F2cos⁡θ52F^2 + 48F^2 \cos \theta = 40F^2 + 24F^2 \cos \theta 48F2cos⁡θ−24F2cos⁡θ=40F2−52F248F^2 \cos \theta - 24F^2 \cos \theta = 40F^2 - 52F^2 24F2cos⁡θ=−12F224F^2 \cos \theta = -12F^2 cos⁡θ=−1224=−12\cos \theta = -\frac{12}{24} = -\frac{1}{2}

For cos⁡θ=−12\cos \theta = -\frac{1}{2}, the angle θ=120∘\theta = 120^\circ.

This is a valid physical angle and is one of the options.

Therefore, the angle between the forces is 120°.