What is 3xY=4w solve for y
in formula transformation

Answer:

The wrong items are;

1) The normal for FN equals the weight Fmg

2) The force of friction, Ff, equals the applied force Fpull

The corrected statements are;

1) The normal force is weight less the vertical component of the applied force Fpull

FN = Fmg - Fpull × sin(θ)

2) The force of friction equals the horizontal component of the applied force Fpull

Ff = Fpull × cos(θ)

Explanation:

The given statement was;

The velocity of the block is constant, so the net force exerted on the block must be zero. Thus, the normal force FN equals the weight Fmg, and the force of friction Ff equals the applied force Fpull

By the equilibrium of forces actin on the system, given that the applied force acts at an angle, θ, with the horizontal, we have;

The normal force is equal to the weight less the vertical component of the applied force;

That is we have, FN = Fmg - Fpull × sin(θ)

The friction force similarly, is equal to the horizontal component of the applied force;

Ff = Fpull × cos(θ)

The wrong items are therefore as follows;

1) The normal for FN equals the weight Fmg

1 i) The normal force is weight less the vertical component of the applied force Fpull

FN = Fmg - Fpull × sin(θ)

2) The force of friction, Ff, equals the applied force Fpull

2 i) The force of friction equals the horizontal component of the applied force Fpull

Ff = Fpull × cos(θ).

Final answer:

While Blake's statement about the normal force is correct, his claim about the applied force and friction force is partially accurate. In reality,the horizontal component of the applied force should equate to the friction force for the block to maintain a constant velocity.

Explanation:

Blake's claim that the normal force FN equals the weight Fmg is correct as these forces balance each other in the vertical direction. However, his claim that the force of friction Ff equals the applied force Fpull is only partially accurate. In reality, the horizontal component of Fpull (i.e., Fpull * cos(θ)) should equate to the friction force Ff, to maintain the constant velocity (the block is not accelerating). The vertical component of Fpull (i.e., Fpull * sin(θ)) reduces the effective weight of the block and thereby, the normal force.

To correct Blake's claim, the normal force FN is equal to the weight of the block minus the vertical component of the applied force, and the applied force's horizontal component equals the friction force. Hence, this is the correct solution considering both vertical and horizontal components of forces.

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If the input machine produces an output of 80 Newtons, then the mechanical advantage is that it produces work. If the required output must be 80 Newtons, then the input force is desirable having 100 percent production of force even though it requires a large amount of input to produce 80 Newtons of force.

its is turning into electrical energy