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3 years ago

In the given arrangement, the normal force applied by block on the ground is

Physics

1 answer:

barxatty [35]3 years ago

7 0

Answer:

The normal force applied by block on the ground is (mg - F cosФ)

Explanation:

Lets explain how to solve the problem

At first you must distribute the force F into two components

Vertical component which is F cosФ

Horizontal component which is F sinФ

The block is in equilibrium, that means sum of forces acting on the

block is zero

So the upward forces equal the downward forces

Normal reaction force R applied by block on the ground and the

vertical component of F both are upward forces

The weight of the block is downward force

The normal reaction force R plus the vertical component of F is

equal to the weight

R + F cosФ = W

W = mg, where g is acceleration of gravity and m is the mass of

the block

R + F cosФ = mg

Subtract F cosФ from both sides

R = mg - F cosФ

The normal force applied by block on the ground is (mg - F cosФ)

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⇒(1.089)(T - 23.8) = (4.186)(79.8 - T)

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⇒(149.193)(T - 23.8) = (941.85)(79.8 - T)

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⇒1091.043 T = 78710.4234

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137(1.089)(T - 23.8) + 11.7(4.186)(T - 5.2)= 225(4.186)(79.8 - T)

⇒(149.193)(T - 23.8) + 48.9762(T - 5.2) = (941.85)(79.8 - T)

⇒149.193 T - 3550.7934 + 48.9762 T - 254.67624= 75159.63 - 941.85 T

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⇒1091.043 T + 48.9762 T = 78710.4234 + 254.67624

⇒1140.0192 T = 78965.09964

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3 years ago

In the given arrangement, the normal force applied by block on the ground is​

In the given arrangement, the normal force applied by block on the ground is