Question

A mass of 10 lg is suspended by a rope of length 4m,
from the ceiling force is applied horizontal o the midpoint of the rope such that the top half ope makes an angle of 45degree with the the vertical. ​

Answer 1

m=10kg

l=4meter

In ΔABC,

∠A+∠B+∠C=180∘

⇒45+90+∠C=180∘

⇒∠C=45∘

Ref. image III

System is in equilibrium

⇒∑F y​=may​=0

⇒T1​sin45∘−mg=0

⇒2

​T1​​=mg⇒T1​=mg √2  __(I)

​

∑Fx​=max​=0

⇒F−T1​cos45∘=0

⇒F=T1/√ 2 = mg  (using eq. (I))

=10×10=100N

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Answer 2

Correct Question:

A mass of 10 Kg is suspended by a rope of length 4m,  from the ceiling force is applied horizontal o the midpoint of the rope such that the top half of the rope makes an angle of 45degree with the the vertical. ​Find the value of Force?

Answer:

• The Force applied is 100 N

Explanation:

Given,

Mass of block (m) = 10 Kg

Angle of inclination (θ) = 45°

Firstly, refer the attachment for diagrams.

So, after resolving the forces we get that sin component of tension equals with weight of the block and cos component balances the Force applied.

As we know that system are in equilibrium, therefore they can individually be resolved and then computed to find the value of force.

Case - 1

⇒ T sinθ = mg

⇒ T sin45° = mg

⇒ T × 1/√(2) = 10 × 10

⇒ T/√(2) = 100

⇒ T = 100 √2 N     _____[1]

Case - 2

⇒ T cosθ = F

⇒ T × cos 45° = F

⇒ T × 1/√(2) = F

Substituting the value of T from equation- (1)

⇒ 100√(2) × 1/√(2) = F

⇒ F = 100 N

∴ The Force applied on the midpoint is 100 N.