Math, asked by amrithpradeep2000, 11 months ago

A block of weight 400N resting on a rough horizontal floor supports a block B weighing 200N. The blocks are connected to a string passing over a smooth pulley as shown in figure. Determine the least horizontal force P to be applied to the block A so as to just move it towards right. Coefficient of friction is 0.25​

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Answered by AditiHegde
4

Given:

A block of weight 400N resting on a rough horizontal floor supports a block B weighing 200N. The blocks are connected to a string passing over a smooth pulley.

Coefficient of friction is 0.25​.

To find:

The least horizontal force P to be applied to the block A so as to just move it towards right.

Solution:

Let m1 = 400 N and m2 = 200 N

Tension T and the opposite force P.

Using Newton's second law, we have,

∑ Fx = m1 × ax

T - 0.25N1 = 0

∑ Fy = m1 × ay

N1 - 200N = 0

N1 = 200N

T - 0.25N1 = 0

T = 0.25N1

T = 0.25 × 200

T = 50N

∑ Fx = m2 × ax

T - P + 0.25N1 + 0.25N2 = 0

∑ Fy = m2 × ay

N2 - 400N - N1 = 0

N2 = 400N + N1

N2 = 400N + 200N

N2 = 600N

T - P + 0.25N1 + 0.25N2 = 0

50 - P + 0.25 × 200 + 0.25 × 600 = 0

50 - P + 50 + 150 = 0

P = 250N

Answered by Fatimakincsem
1

The least horizontal force is P = 250 N

Step-by-step explanation:

Let m1 = 400 N and m2 = 200 N

Tension T and the opposite force P.

Using Newton's second law, we have,

∑ Fx = m1 × ax

T - 0.25 N1 = 0

∑ Fy = m1 × ay

N1 - 200 N = 0

⇒ N1 = 200 N

T - 0.25N1 = 0

T = 0.25 N1

T = 0.25 × 200

⇒ T = 50 N

∑ Fx = m2 × ax

T - P + 0.25 N1 + 0.25 N2 = 0

∑ Fy = m2 × ay

N2 - 400 N - N1 = 0

N2 = 400 N + N1

N2 = 400 N + 200 N

⇒ N2 = 600 N

T - P + 0.25 N1 + 0.25 N2 = 0

50 - P + 0.25 × 200 + 0.25 × 600 = 0

50 - P + 50 + 150 = 0

⇒ P = 250 N

The least horizontal force is P = 250 N

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