Physics, 05.10.2019 01:10 ketabug50

Awedge with an inclination of angle θ rests next to a wall. a block of mass m is sliding down the plane. there is no friction between the wedge and the block or between the wedge and the horizontal surface.
find the magnitude, fnet, of the sum of all forces acting on the block.
express fnet in terms of θ and m, along with any necessary constants.
find the magnitude, fww, of the force that the wall exerts on the wedge.
express fww in terms of θ and m, along with any necessary constants.

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Answer from: TGJuuzouSuzuya

The magnitude of the net force acting on the bock is (ma - mgsinθ).

The magnitude of the force that the wall exerts on the wedge is mg(cosθ) (sinθ).

The given parameters;

angle of inclination, = θ

The normal force on the block due to the block's weight is calculated as follows;

$F_n = mgcos(\theta)$

The net horizontal force on the bock is calculated as;

$mgsin(\theta) = ma\\\\\Sigma F_x = ma - mgsin(\theta)$

$\Sigma F_{net} = ma - mgsin(\theta)$

The magnitude of the force the wall exerts on the wedge is equal to the force the wedge exerts on the wall.

The weight of wedge = Fₙ = mgcos(θ)

The net horizontal force on the wall due to the weight of the wedge is calculated as follows;

$\Sigma F_x = 0\\\\F_n sin(\theta) - ma = 0\\\\F_n sin(\theta) - m(0) = 0 \\\\F_nsin(\theta) = 0\\\\\Sigma F_x = F_n sin(\theta)\\\\\Sigma F_x = mg\ cos(\theta)sin(\theta)\\\\\Sigma F_w_w = mg\ cos(\theta)sin(\theta)$

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Answer from: Itsjazzyyy001

The net force on the block F(net) = mgsinθ).

Fw =mg(cosθ)(sinθ)

Explanation:

(a)

Here, m is the mass of the block, n is the normal force, \thetaθ is the wedge angle, and Fw is the force exerted by the wall on the wedge.

Since the block sliding down, the net force on the block is along the plane of the wedge that is equal to horizontal component of weight of the block.

F(net) = mgsinθ

The net force on the block F(net) = mgsinθ).

The direction of motion of the block is along the direction of net force acting on the block. Since there is no frictional force between the wedge and block, the only force acting on the block along the direction of motion is mgsinθ.

(b)

From the free body diagram, the normal force n is equal to mgcosθ .

n=mgcosθ

The horizontal component of normal force on the block is equal to force

Fw=n*sin(θ) that exerted by the wall on the wedge.

Substitute mgcosθ for n in the above equation;

Fw =mg(cosθ)(sinθ)

Since, there is no friction between the wedge and the wall, there is component force acting on the wall to restrict the motion of the wedge on the surface and that force is arises from the horizontal component for normal force on the block.

Answer from: Quest

The rearrangement of particles in a physical change. hope this

Answer from: Quest

it would be c. predictability

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