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3 years ago

A box is at rest on a ramp at an incline of 22°. The normal force on the box is 538 N.

Physics

2 answers:

fomenos3 years ago

8 0

Answer: 580 N

Refer to attached figure.

The angle of inclination is 22 degrees

weight (gravitational force) acts downwards.

Normal force is a contact force which acts perpendicular to the point of contact.

The horizontal component (mg cos 22 ) balances the normal force and the vertical component balances the frictional force.

Gravitational force on an object = mg

The normal force $N= mg cos 22$

$\Rightarrow mg =\frac{N}{cos22}=\frac{538 N}{0.927}=580 N$

liubo4ka [24]3 years ago

7 0

Answer

580

Explanation

The gravitational force acts perpendicularly downwards (towards the center of the earth).

The normal force of the box is perpendicular to the slanting ramp. This force we are given 538 N.

The gravitational force is the component of this 538 N.

∴ Cos22° = 538/Y Where y is the gravitational force.

Cos22° = 538/Y

Y× Cos22° = 538

Y = 538/Cos22°

= 580.25

Answer to the nearest whole number = 580.

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2 years ago

What is the magnitude of the torque that the axle must apply to prevent the disk from rotating?

mihalych1998 [28]

The required torque at the axle, is given by the difference between the

moments of the applied forces.

The torque required is 19.62 N·m counterclockwise

Reasons:

The given parameters are;

Mass of the disk, m = 5.0 kg

Location of the axle = Half the radius of the disk

Diameter of the disk, D = 40 cm = 0.4 m

Applied mass, 0.1 m from the axle = 15 kg

Applied mass, 0.3 m from the axle = 10 kg

Required:

Magnitude of torque at the axle that prevent the disk from rotating

Solution:

Torque needed = Clockwise moment - Counterclockwise moment

Clockwise moment = (10 kg × 0.3 m + 5 kg × 0.1 m) × 9.81 m/s² = 34.335 N·m

Counterclockwise moment = 15 kg × 0.1 m × 9.81 m/s² = 14.715 N·m

τ + Counterclockwise moment = Clockwise moment

τ + 14.715 N·m = 34.335 N·m

Torque required, τ = 34.335 N·m - 14.715 N·m = 19.62 N·m

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2 years ago

An object, when pushed with a net force F, has an acceleration of 4 m/s . Now twice the force is applied to an object that has f

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Answer:

B. 2 m/s

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Explanation:

A force is applied on a mass m whose acceleration is 4 m/s

Force = mass × acceleration

a = F/m = 4 m/s

4 m/s = F/m

F = 4 m/s (m)

If Force of 2F is applied on a mass of 4m ; it acceleration is as follows:

2F/4 m = F/ 2m

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mass $m_1$ = 30 kg

mass $m_2$ = 50 kg

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$m_{1} a _ \ {net} }= T - mg sin \theta - F$ ---- equation (2)

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From equation , replace a with 4.05

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T = 500 -202.5

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