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Sauron [17]
3 years ago
5

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:

<em>Lets explain how to solve the problem</em>

At first you must distribute the force F into two components

<em>Vertical</em> component which is <em>F cosФ</em>

<em>Horizontal</em> component which is <em>F sinФ</em>

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

<em>R + F cosФ = W</em>

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

the block

<em>R + F cosФ = mg</em>

Subtract F cosФ from both sides

<em>R = mg - F cosФ</em>

<em></em>

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

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