for the third part since frictional force is more than then force, shouldn't particle B be at rest? How can the particle move with negative acceleration? Please explain i have attached both question and answer

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Chapter1: Units, Trigonometry. And Vectors
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for the third part since frictional force is more than then force, shouldn't particle B be at rest? How can the particle move with negative acceleration? Please explain i have attached both question and answer

(i) [1.2 = mg cosa]
Mass is 0.125 kg
(ii) [-mg sina – F= ma]
- 0.125 x 10 x 0.28 – 0.4 = 0.125a
a = -6 > deceleration is 6 ms
-2
(iii)
µR > mg sina → particle remains at rest
Transcribed Image Text:(i) [1.2 = mg cosa] Mass is 0.125 kg (ii) [-mg sina – F= ma] - 0.125 x 10 x 0.28 – 0.4 = 0.125a a = -6 > deceleration is 6 ms -2 (iii) µR > mg sina → particle remains at rest
A particle moves up a line of greatest slope of a rough plane inclined at an angle a to the horizontal,
where cos a = 0.96 and sin a = 0.28.
4
(i) Given that the normal component of the contact force acting on the particle has magnitude 1.2 N,
find the mass of the particle.
(ii) Given also that the frictional component of the contact force acting on the particle has magnitude
0.4 N, find the deceleration of the particle.
The particle comes to rest on reaching the point X.
(iii) Determine whether the particle remains at X or whether it starts to move down the plane.
Transcribed Image Text:A particle moves up a line of greatest slope of a rough plane inclined at an angle a to the horizontal, where cos a = 0.96 and sin a = 0.28. 4 (i) Given that the normal component of the contact force acting on the particle has magnitude 1.2 N, find the mass of the particle. (ii) Given also that the frictional component of the contact force acting on the particle has magnitude 0.4 N, find the deceleration of the particle. The particle comes to rest on reaching the point X. (iii) Determine whether the particle remains at X or whether it starts to move down the plane.
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