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1 month 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:

ValentinkaMS [3.4K]1 month ago

3 0

The full question reads;

Jason is employed at a moving company. A wooden crate weighing 75 kg is positioned on the wooden ramp of his truck, inclined at an angle of 11°.

What is the force magnitude, directed parallel to the ramp, that he needs to apply to initiate the upward movement of the crate?

Answer:

F = 501.5 N

Explanation:

We have the following information;

Mass of the wooden crate; m = 75 kg

Incline angle; θ = 11°

To move the wooden crate up, we must consider that friction is acting in the opposite direction of the movement along the inclined surface. Therefore, the force required can be expressed by;

F = mgsin θ + μmg cos θ

Using online resources, the 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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A block of mass 3m is placed on a frictionless horizontal surface, and a second block of mass m is placed on top of the first bl

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According to Newton's second law, provided F is directed horizontally,

• the net horizontal force acting on the larger block is

F - µmg = 3mA

where µmg represents the friction experienced by the larger block from contact with the smaller block, µ is the static friction coefficient between both blocks, and A indicates the acceleration of the block;

• the net vertical force on the larger block is

4mg - 3mg - mg = 0

where 4mg denotes the magnitude of the normal force exerted by the surface on the combined mass of both blocks, 3mg corresponds to the weight of the larger block, and mg indicates the weight of the smaller block;

• the net horizontal force acting on the smaller block is

µmg = ma

where µmg again signifies the friction between both blocks; however, it is important to note that this force aligns in the same direction as F. It is the sole force influencing the smaller block in the horizontal direction, thus (b) static friction causes the smaller block's acceleration;

• the net vertical force on the smaller block is

mg - mg = 0

where mg represents the force of both the normal force from the larger block pushing up against the smaller one, and the weight of the smaller block.

(You should be able to create your own free-body diagrams based on the forces discussed.)

(c) Solve the equations stated above to find A and a:

A = (F - µmg) / (3m)

a = µg

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1 month ago

Jeff's body contains about 5.46 L of blood that has a density of 1060 kg/m3. Approximately 45.0% (by mass) of the blood is cells

ValentinkaMS [3465]

Answer:

a) Blood mass is $m= 5.7876kg$

b) The count of blood cells is $N_t=1.04*10^{13}$

Explanation:

From the problem statement, we learn that

The blood volume is $V_b = 5.46 \ L = \frac{5.46}{1000} = 0.00546m^3$

The density of blood is $\rho_b = 1060 kg/m^3$

% of blood which consists of cells is = 45.0%

the % of blood that is plasma is = 55.0%

density of blood cells is $\rho_d = 1125kg/m^3$

% of cells that are white is = 1%

% of cells that are red is = 99%

The red blood cell diameter is = $7.5 \mu m = 7.5*10^{-6}m$

The red blood cell radius is $= \frac{7.5*10^{-6}}{2} = 3.75*10^{-6}m$

The mass is generally represented mathematically as

$m = \rho_b * V_b$

Substituting values

$m = 1060 * 0.00546$

$m= 5.7876kg$

Cell mass is $m_c$ = 45% of m

= 0.45 * 5,7876

= 2.60442 kg

The volume of cells is $V_c$ = $\frac{m_c}{\rho_d}$

$= \frac{2.60442}{1125}$

$= 2.315 *10^{-3} m^3$

The white blood cells volume is $V_w$ = 1% of the cells volume

$= \frac{1}{100} * 2.315*10^{-3}$

$= 2.315*10^{-5}m^3$

The volume of a single cell is $V_s = 4 \pi r^3$

$= 4*(3.142) * (3.75*10^{-6})^3$

$= 2.21*10^{-16}m^3$

The red blood cells volume is $V_r = V_c - V_w$

$=2.315*10^{-3} - 2.315*10^{-5}$

$= 2.29*10^{-3}m^3$

The total red blood cell count is $= \frac{V_r}{V_s}$

$= \frac{2.29 *10^{-3}}{2.21*10^{-16}}$

$= 1.037*10^{13}$

The total white blood cell count is $=\frac{V_w}{V_s}$

$= \frac{2.315 * 10^{-5}}{2.21*10^{-16}}$

$= 1.04*10^{11}$

The overall number of blood cells is $N_t= 1.037*10^{13} + 1.04*10^{11}$

$N_t=1.04*10^{13}$

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2 months ago

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An end of a light wire rod is bent into a hoop of radius r. The straight part of the rod has length l; a ball of mass M is attac

ValentinkaMS [3465]

Answer:

$arcsin(\frac{R\mu}{(R+l)\sqrt{\mu^2+1}})$

Explanation:

By applying the Law of Sines,

$sin \theta = \frac{sin \phi R}{ l + R}$

Based on Newton's Law,

$mg = N\sqrt{\mu^2+1}$

And the final equation also derived from Newton's Law,

$\mu N = mgsin\phi$

Then by consolidating all the equations together,

$\mu N = mgsin\phi = N\sqrt{\mu^2+1}sin\phi\\$

$sin\theta = \frac{\mu R}{ (l + R)\sqrt{\mu^2+1}}$

Thus,

$\theta = arcsin(\frac{R\mu}{(R+l)\sqrt{\mu^2+1}})$

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The gravitational force between Royce and Earth would double by the age of 16. Newton’s law of universal gravitation states that the gravitational force is proportional to the masses involved and inversely proportional to the square of the distance between them. At age 10, Royce weighed 30 kg, and by age 16, he weighed 60 kg. Since his mass doubled from 10 to 16 years, this results in a corresponding doubling of the gravitational force.

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