Fig. shows a uniform rod of length 30 cm having a mass of
3.0 kg. The strings shown in the figure are pulled by constant
forces of 20 N and 32 N. All the surfaces are smooth and the
strings and pulleys are light. The force exerted by 20 cm part
of the rod on the 10 cm part is
(a) 20 N (b) 24 N
(c) 32 N (d) 52 N
Answers
Explanation:
HEY MATE HERE IS YOUR ANSWER
Mass per unit length= 303 kg/cm=0.1kg/cmMass of 10cm part=m 1 =1kgMass of 20cm part=m 2 =2kgLet F be the force of contact between themFrom free body diagram,F−20−a=0 ......(1)32−F−2a=0 ......(2)⇒32−20−2a−a=0⇒12−3a=0⇒a= 312 =4ms −2 Contact forceF=20+a=20+4=24N∴ force exerted=24N
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Answer:
The force exerted by 20 cm part of the rod on the 10 cm part is 24 N.
Explanation:
Given:-
Mass of rod, m = 3 kg
Length of rod, L = 30 cm
Forces acting, F₁ & F₂ = 20 N & 32 N respectively.
[Refer to the attached image for the visualization of the scenario]
In the given case;
Net Force, F = 32 - 20
F = 12 N
We know that by formula:
Acceleration, a = net pulling force/ net mass
a = F/m
a = 12/ 3
a = 4 m/s²
Now, from the attachment, we have tension forces acting opposite to the parts 20 cm and 10 cm in contact. From this we have:
T - F₁ = ma
T - 20 = 1 × 4
T = 20 + 4
T = 24 N _____(1)
Similarly,
F₂ - T = ma
32 - T = 2 × 4
32 - 8 = T
T = 24 N ______(2)
From (1) and (2), we conclude that:
The force exerted by 20 cm part of the rod on the 10 cm part or 10 cm part of the rod on the 20 cm part is 24 N.
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