block on inclined plane

awepratization

2010-01-22 6:36 am

A 3.5kg block on a smooth inclined plane of angle 30 degrees is connected by a cord over a small frictionless pulley to a 2.5kg block hanging vertically.

(a) When the blocks are released, calculate
(i) the acceleration of the system,
(ii) the tension in the cord.

(b) A force is applied onto one of the blocks to keep the system stationary. Find the minimum force required, if it is appled on
(i) the 3.5kg block,
(ii) the 2.5kg block.

回答 (1)

天同

2010-01-22 5:10 pm

✔ 最佳答案

(a) (i) Net force on the system = 2.5g - 3.5gsin(30) N = 7.5 N
where g is the acceleration due to gravity (= 10 m/s2)
hence, acceleration = force/mass = 7.5/(3.5+2.5) m/s2 = 1.25 m/s2

(ii) Consider the 2.5 kg block, let T be the tension in the cord
T - 2.5g = 2.5 x 1.25
hence, T = (2.5x1.25 +2.5g) N = 28.125 N

(b) (i) Assume the force applied to the 3.5 kg block is parallel to the inclined plane, let the force be F,
At equilibrium, F + 3.5gsin(30) = T'
where T' is the new tension in the cord

But at equilibrium, T' = 2.5g N
hence, F = (2.5g - 3.5gsin(30)) N = 7.5 N

(ii) Let F" be the force, and T" is the tension in the cord
F" + T" = 2.5g
but T" = 3.5gsin(30) N = 17.5 N
hence, F" = (2.5g - 17.5) N = 7.5 N

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