Find the force necessary to start the crate moving, given that the mass of the crate is 32 kg and the coefficient of static friction between the crate and the floor is 0.57. Express your answer using two significant figures.

Answers

Answer 1

Answer:

178.75 N

Explanation:

The force necessary to start moving the crate must be equal to or more than the frictional force (resistive force) acting on the crate but moving in an opposite direction to the frictional force.

So, we find the frictional force, Fr:

Fr = -μmg

Where μ = coefficient of friction

m = mass

g = acceleration due to gravity

The frictional force is negative because it acts against the direction of motion of the crate.

Fr = -0.57 * 32 * 9.8

Fr = - 178.75 N

Hence, the force necessary to move the crate must be at least equal to but opposite in direction to this frictional force.

Therefore, this force is 178.75 N

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Observer 1 rides in a car and drops a ball from rest straight downward, relative to the interior of the car. The car moves horizontally with a constant speed of 3.80 m/s relative to observer 2 standing on the sidewalk.a) What is the speed of the ball 1.00 s after it is released, as measured by observer 2?

b) What is the direction of travel of the ball 1.00 s after it is released, as measured relative to the horizontal by observer 2?

Answers

a) 10.5 m/s

While for observer 1, in motion with the car, the ball falls down straight vertically, according to observer 2, which is at rest, the ball is also moving with a horizontal speed of:

$v_x = 3.80 m/s$

As the ball falls down, it also gains speed along the vertical direction (due to the effect of gravity). The vertical speed is given by

$v_y = u_y + gt$

where

$u_y =0$ is the initial vertical speed

g = 9.8 m/s^2 is the acceleration of gravity

t is the time

Therefore, after t = 1.00 s, the vertical speed is

$v_y = 0 + (9.8)(1.00)=9.8 m/s$

And so the speed of the ball, as observed by observer 2 at rest, is given by the resultant of the horizontal and vertical speed:

$v=√(v_x^2 +v_y^2)=√((3.8)^2+(9.8)^2)=10.5 m/s$

b) $\theta = -68.8^(\circ)$

As we discussed in previous part, according to observer 2 the ball is travelling both horizontally and vertically.

The direction of travel of the ball, according to observer 2, is given by

$\theta = tan^(-1) ((v_y)/(v_x))=tan^(-1) ((-9.8)/(3.8))=-68.8^(\circ)$

We have to understand in which direction is this angle measured. In fact, the car is moving forward, so $v_x$ has forward direction (we can say it is positive if we take forward as positive direction).

Also, the ball is moving downward, so $v_y$ is negative (assuming upward is the positive direction). This means that the direction of the ball is forward-downward, so the angle above is measured as angle below the positive horizontal direction:

$\theta = -68.8^(\circ)$

The knot at the junction is in equilibrium under the influence of four forces acting on it. The F force acts from above on the left at an angle of α with the horizontal. The 5.7 N force acts from above on the right at an angle of 50◦ with the horizontal. The 6.2 N force acts from below on the right at an angle of 44◦ with the horizontal. The 6.7 N force acts from below on the left at an angle of 43◦ with the horizontal.1. What is the magnitude of the force F?
2. What is the angle a of the force F in the figure above?

Answers

(a) The magnitude of the force F acting on the knot is 5.54 N.

(b) The angle α of the force F is 54.4⁰.

The given parameters:

F force at α
5.7 N force at 50⁰
6.2 N force at 44⁰
6.7 N force at 43⁰

The net vertical force on the knot is calculated as follows;

$F_y = Fsin(\alpha) + 5.7 sin(50) - 6.2 sin(44) - 6.7 sin(43)\n\nF_y = F sin(\alpha) -4.51\n\nFsin(\alpha) = 4.51$

The net horizontal force on the knot is calculated as follows;

$F_x = -F cos(\alpha) + 5.7 cos(50) + 6.2cos(44) - 6.7cos(43)\n\nF_x = -Fcos(\alpha) + 3.22\n\nFcos(\alpha) = 3.22$

From the trig identity;

$sin^2 \theta + cos^ 2 \theta = 1\n\n$

$(Fsin(\alpha))^2 + (Fcos(\alpha))^2 = (4.51)^2 + (3.22)^2\n\nF^2(sin^ 2\alpha + cos^2 \alpha) = 30.71\n\nF^2(1) = 30.71\n\nF = √(30.71) \n\nF = 5.54 \ N$

The angle α of the force F is calculated as follows;

$Fsin(\alpha) = 4.51\n\nsin(\alpha) = (4.51)/(F) \n\nsin(\alpha ) = (4.51)/(5.54) \n\nsin(\alpha ) = 0.814\n\n\alpha = sin^(-1)(0.814)\n\n\alpha = 54.5 \ ^0$

Find the image uploaded for the complete question.

Learn more about net force here:brainly.com/question/12582625

The knot is in equilbrium, so there is no net force acting on it. Starting with the unknown force and going clockwise, denote each force by F₁, F₂, F₃, and F₄, respectively. We have

F₁ + F₂ + F₃ + F₄ = 0

Decomposing each force into horizontal and vertical components, we have

F cos(180º - α) + (5.7 N) cos(50º) + (6.2 N) cos(-44º) + (6.7 N) cos(-137º) = 0

F sin(180º - α) + (5.7 N) sin(50º) + (6.2 N) sin(-44º) + (6.7 N) sin(-137º) = 0

Recall that cos(180º - x) = - cos(x) and sin(180º - x) = sin(x), so these equations reduce to

F cos(α) ≈ - 3.22 N

F sin(α) ≈ 4.51 N

(1) Recall that for all x, sin²(x) + cos²(x) = 1. Use this identity to solve for F :

(F cos(α))² + (F sin(α))² = F ² ≈ 30.73 N² → F ≈ 5.5 N

(2) Use the definition of tangent to solve for α :

tan(α) = sin(α) / cos(α) ≈ 1.399 → α ≈ 126º

or about 54º from the horizontal from above on the left of the knot.

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Swinging a tennis racket against a ball is an example of a third class lever.
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Please select the best answer from the choices provided.
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Answers

Final answer:

Swinging a tennis racket against a ball as a third class lever in physics.

Explanation:

Swinging a Tennis Racket as a Third Class Lever

A tennis racket swinging against a ball is indeed an example of a third class lever in physics. In a third class lever, the effort is situated between the fulcrum and the load. In this case, the effort is provided by the player's hand gripping the racket handle, the fulcrum is the wrist joint, and the load is the ball being struck by the racket.

When a player swings the racket, the force applied by the player's hand exerts an effort on the handle of the racket. This causes the racket to rotate about the wrist joint acting as the fulcrum. The ball serves as the load, receiving the force and accelerating in the opposite direction.

Learn more about Third class lever here:

brainly.com/question/4532561

A block, initially at rest, has a mass m and sits on aplane inclined at angle (theta). It slides a distance d before hitting a spring and compresses the spring by a maximum distance of xf. If the coefficient of kinetic friction between the plane and block is uk, then what is the force constant of the spring?

Answers

Answer: k = ma + uk×mgcosθ/ xf

Explanation: The body is placed on a frictionless inclined ramp.

The weight of the object has 2 components, horizontal component (mgsinθ) and vertical component (mgcosθ).

The horizontal component of weight is responsible for making tje object slide down the plane even with no applied force.

So from newton's second law of motion

mgsinθ - uk×R = ma

Where uk = coefficient of kinetic friction.

R = normal reaction = mgcosθ

mgsinθ - uk×mgcosθ = ma

mgsinθ = ma + uk×mgcosθ

mgsinθ is the applied force in this case. This applied force compresses a spring.

According to hooke's law,

F =ke

Where F = ma + uk×mgcosθ, e =xf

F = applied force , e = extension and k = spring constant.

k = F/e

k = ma + uk×mgcosθ/ xf

Your classmate’s mass is 63 kg and the table weighs 500 N. Calculate the normal force on the table by the floor. Show your work!

Answers

Answer:

$F_N=1234.8N$

Explanation:

Hello.

In this case, since the normal force is opposite to the total present weight, we can compute it by considering the mass of the classmate with the gravity to compute its weight, and the weight of the table:

$F_N=63kg*9.8m/s^2+500N\n\nF_N=617.4N+500N\n\nF_N=1234.8N$

Best regards.

What operation do you apply to the position function of a particle to compute the particles velocity

Answers

Answer:

the derivative with respect to time

Explanation:

This is an exercise in kinematics, where the velocity is defined as a function of the position of a body of the form

v = dx/dt

where v is the velocity of the body, x is the position that we assume is a continuous and differentiable function.

The function written in the equation is the derivative with respect to time

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