mechanics question (1 Viewer)

[maths lover]

not sure about questions which relate to the conical pendulum but use a pole instead of a string. i remember my teacher saying something about poles in that they can both pull and push, however i am not sure as whether it for this reason or not that they have a reaction force on top of the regular normal force. also is this reaction force always perpendicular to the normal force?. an example of this question is q16 on page 234 of the Cambridge textbook. any help is appreciated thanks.

edit: hopefully you understand what i am saying, if not i can clarify.

 

[Drongoski]

Here we treat the 2 rods as if they are 2 strings. ABC constitutes an isosceles triangle with AB and BC making an angle of 30 degrees with the vertical. Let the tension forces for AB and BC be T₁ and T₂ resp.

For forces on 2 kg mass at B:

Centripetal force = "mw²r" = 2 x 6² x 1/4 Newtons = 18 N [provided by the horizontal components of T₁ and T₂ at B]

.: T₁ sin 30 + T₂ sin 30= 18 ==> T₁ + T₂ = 36 N . . . . [1]

For vertical forces at B:

T₁ cos 30 = 2g + T₂ cos 30 ==> T₁ - T₂ = 4g/sqrt(3) = 40/sqrt(3) . . . . [2]

At C, with R the vertical normal reaction R on the 1 kg ring:

R + T₂ cos 30 = 1 g = 10 N . . . . [3]

Solving [1] & [2], we get:

T₁ = 18 + 2g/sqrt(3) = 18 + 20/sqrt(3) N

T₂ = 18 - 2g/sqrt(3) = 18 - 20/sqrt(3) N

Substitute this value of T₂ into [3] we get:

R = g - sqrt(3)/2 T₂ = 20 - 9 sqrt(3) N

 

[maths lover]

Here we treat the 2 rods as if they are 2 strings. ABC constitutes an isosceles triangle with AB and BC making an angle of 30 degrees with the vertical. Let the tension forces for AB and BC be T₁ and T₂ resp.

For forces on 2 kg mass at B:

Centripetal force = "mw²r" = 2 x 6² x 1/4 Newtons = 18 N[provided by the horizontal components of T₁ and T₂ at B]

.: T₁ sin 30 + T₂ sin 30= 18 ==> T₁ + T₂ = 36 N . . . . [1]

For vertical forces at B:

T₁ cos 30 = 2g + T₂ cos 30 ==> T₁ - T₂ = 4g/sqrt(3) = 40/sqrt(3) . . . . [2]

At C, with R the vertical normal reaction R on the 1 kg ring:

R + T₂ cos 30 = 1 g = 10 N . . . . [3]

Solving [1] & [2], we get:

T₁ = 18 + 2g/sqrt(3) = 18 + 20/sqrt(3) N

T₂ = 18 - 2g/sqrt(3) = 18 - 20/sqrt(3) N

Substitute this value of T₂ into [3] we get:

R = g - sqrt(3)/2 T₂ = 20 - 9 sqrt(3) N

thanks heaps for answering the question, but could you explain to me the reason as to why there is a reaction force perpendicular to the normal force on c?.

 

[Drongoski]

There is a net vertical force acting down on the ledge and therefore a corresponding reaction in the opposite direction (up). There is also a horizontal tug on the vertical rod at C which will therefore exert an equal horizontal normal reaction to that.

 

[maths lover]

There is a net vertical force acting down on the ledge and therefore a corresponding reaction in the opposite direction (up). There is also a horizontal tug on the vertical rod at C which will therefore exert an equal horizontal normal reaction to that.

i cant thank you enough you just clarified allot. im guessing the tug to the right is a result of the tension from the pole?.

 

[Drongoski]

Yep.

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