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2 years ago

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What is the magnitude of the force, directed parallel to the ramp, that he needs to exert on the crate to get it to start moving

up the ramp

1 answer:

lions [1.4K]2 years ago

3 0

Complete question is;

Jason works for a moving company. A 75 kg wooden crate is sitting on the wooden ramp of his truck; the ramp is angled at 11°.

What is the magnitude of the force, directed parallel to the ramp, that he needs to exert on the crate to get it to start moving UP the ramp?

Answer:

F = 501.5 N

Explanation:

We are given;

Mass of wooden crate; m = 75 kg

Angle of ramp; θ = 11°

Now, for the wooden crate to slide upwards, it means that the force of friction would be acting in an opposite to the slide along the inclined plane. Thus, the force will be given by;

F = mgsin θ + μmg cos θ

From online values, coefficient of friction between wooden surfaces is μ = 0.5

Thus;

F = (75 × 9.81 × sin 11) + (0.5 × 75 × 9.81 × cos 11)

F = 501.5 N

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2 years ago

Rod AB is held in place by the cord AC. Knowing that the tension in the cord is 1350 N and that c 5 360 mm, determine the moment

riadik2000 [5.3K]

Answer:

291.598 N-m

291.6 N-m

Explanation:

Let's first take a look at the free bodily diagrammatic representation.

The first diagram will aid us in answering question (a), so as the second diagram will facilitate effective understanding when solving for question (b).

Let's first determine our angle θ from the diagram

To find angle θ ; we have :

tan θ = $\frac{360+240}{450}$

tan θ = $\frac{600}{450}$

tan θ = 1.333

θ = tan⁻¹ (1.333)

θ = 53.13°

Now, to determine the moment about B of the force exerted by the cord at point A by resolving that force into horizontal and vertical components applied at point A.

We have:

$M__B}=(Fcos \theta *240)-(Fsin \theta *450)$

where Force(F) = Force in the cord AC = 1350 N and θ = 53.13° ; we have:

$M__B}=(1350&cos 53.13^0 *240)-(1350sin 53.13^0 *450)$

$M__B}= 194400.463-485999.348$

$M__B}=-291598.885 N-mm\\$

$M__B}=-291.598 N-m$

Since the negative sign illustrates just the clockwise movement ; then the moment about B of the force exerted by the cord at point A by resolving that force into horizontal and vertical components applied at point A = 291.598 N-m

b) From the second diagram, taking the moment at point B $(M__B})$ ,

we have:

$M__B}=-Fcos \theta *360 - Fsin \theta * 0$

$M__B}=-Fcos \theta *360 - 0$

$M__B}=-Fcos \theta *360$

where Force(F) = 1350 N and θ = 53.13° ; we have:

$M__B}= -1350*cos53.13^0*360$

$M__B}= -291600 N-mm$

$M__B}= -291.6 N-m$

Since the negative sign illustrates just the clockwise movement ; then the moment about B of the force exerted by the cord at point A by resolving that force into horizontal and vertical components applied at point C = 291.6 N-m

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2 years ago

A lawn sprinkler sprays water 8 feet at full pressure as it rotates 360 degrees. if the water pressure is reduced by 50%, what i

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The area of the sprinkles can be determined through the area of a circle that is pi * r^2 in which the given dimensions above are the radii, r. The second scenarios radius is only half of the original, that is 4 ft. In this case, we can compute the area of the second again. We calculate next the difference of two areas of circles.

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2 years ago

A child's toy consists of a m = 36 g monkey suspended from a spring of negligible mass and spring constant k. When the toy monke

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Answer:

Part A - 3N/m

Part B - see attachment

Part C - 4.9 × 10-³J

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Part D

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2 years ago

A tennis ball of mass m=0.060 kg and speed v=25 m/s strikes a wall at a 45 angle and rebounds with the same velocity at 45°. Wha

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To solve this problem we will apply the concepts related to the Impulse which can be defined as the product between mass and the total change in velocity. That is to say

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Here,

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As we can see there are two types of velocity at the moment the object makes the impact,

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That is to say,

$v_i = vcos\theta$

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Therefore the impulse would be as

$p = 0.060(-17.68-17.68)$

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The negative sign indicates that the pulse is in the opposite direction of the initial velocity.

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2 years ago