Physics Laws of Motion

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2 weeks ago

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R
Raj Pandey

Contributor-Level 9

Contact force at C 
a=5m/sec2 =20N

Contact force at   Contact force at

A=6*5=30B=5*5=25N

New answer posted

2 weeks ago

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R
Raj Pandey

Contributor-Level 9

F cos 30o component is responsible for motion and tension

F cos 30? = (3+12+15)a

a = 3 F 6 0

T21= 3F360 ]

  T 1 1 = 1 5 F 3 6 0

T 2 T 1 = 5 1

T 1 T 2 = 5 1

T 1 : T 2 = 5 : 1

 

New answer posted

2 weeks ago

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V
Vishal Baghel

Contributor-Level 10

f ? m g

? ? m r ? 2 ? m g

? ? = g r ? 2 = 1 0 1 * 1 0 0 = 0 . 1

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R
Raj Pandey

Contributor-Level 9

Let is the angle made by the wire with the vertical.

Here,  .

 

 

 

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A
alok kumar singh

Contributor-Level 10

Balancing torque about COM

T A l 2 = T B l 6

? T A T B = 1 3

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V
Vishal Baghel

Contributor-Level 10

As per Newton third law, the horse exerts a force on the ground, the ground will also exert a force on the horse.

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V
Vishal Baghel

Contributor-Level 10

l = m (2v) = 1 * 2 * 15 = 30 kg m s-1

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V
Vishal Baghel

Contributor-Level 10

f l = μ N = 0 . 6 * m g c o s θ f l = 0 . 6 * 2 * 1 0 * 4 5 = 9 . 6 N

Component of weight along incline.

= m g s i n θ = 2 * 1 0 * 3 5 = 1 2 N

m g s i n θ > f l s o T = 2 . 4 N

New answer posted

2 weeks ago

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V
Vishal Baghel

Contributor-Level 10

K . E . t 2

1 2 m v 2 t 2 v t d v d t = c o n s t a n t F = m a = m d v d t = c o n s t a n t

New answer posted

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A
alok kumar singh

Contributor-Level 10

Kindly go throug the solution

 

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