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irga5000 [103]
3 years ago
12

Because the block is not moving, the sum of the x components of the forces acting on the block must be zero. Find an expression

for the sum of the x components of the forces acting on the block, using coordinate system b.
Physics
1 answer:
Viktor [21]3 years ago
7 0

Answer:

F_{xtotal} = F_{x} - \mu N

Explanation:

Let the total force be F

The force that produces motion in the x direction is given as, F_{x}

Let the frictional force be given as:

F_{fric} = \mu N

Thus, the net force is given as:

F_{xnet}  = F_{x} - \mu N

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A mother and daughter are on a seesaw in the park. How far from the center must the 160.9lb mother sit in order to balance the 6
Soloha48 [4]

Answer:

The mother has to sit 2.17 ft from the center on the other side of the seesaw.

Explanation:

We are trying to find the sum of torques given by the weights of mother and daughter to be zero.

If the torque of the daughter on one side of the pivoting point is given by:

5.5 ft x 63.5 lb x g = 349.25 g ft lb

we need that the absolute value of the torque exerted by the mom (160.9 lb) to be the same in magnitude (and of course opposite direction). So we assume that "d" is the distance at which the mother locates to make this torque equal in magnitude to her daughter's torque:

d x 160.9 lb x g = 349.25 g ft lb

d = 2.17 ft

3 0
3 years ago
A horizontal force, F1 = 65 N, and a force, F2 = 12.4 N acting at an angle of θ to the horizontal, are applied to a block of mas
Nezavi [6.7K]

Answer:

(a) FN = 24.18 N

(b) a = 22.87 m/s²

Explanation:

Newton's second law of the  block:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Forces acting on the box

We define the x-axis in the direction parallel to the movement of the block on the surface   and the y-axis in the direction perpendicular to it.

F₁ : Horizontal force

F₂ : acting at an angle of θ to the horizontal,

W: Weight of the block  : In vertical direction

FN : Normal force : perpendicular to the direction the surface

fk : Friction force: parallel to the direction to the surface

Known data

m =3.1 kg : mass of the  block

F₁ = 65 N,  horizontal force

F₂ = 12.4 N acting at an angle of θ to the horizontal

θ = 30° angle θ of F₂ with respect to the horizontal

μk = 0.2 : coefficient of kinetic friction between the block and the surface

g = 9.8 m/s² : acceleration due to gravity

Calculated of the weight  of the block

W= m*g  = (3.1 kg)*(9.8 m/s²) = 30.38 N

x-y F₂ components

F₂x = F₂cos θ= (12.4)*cos(30)° = 10.74 N

F₂y = F₂sin θ= (12.4)*sin(30)° = 6.2 N

a)Calculated of the Normal force  (FN)

We apply the formula (1)

∑Fy = m*ay    ay = 0

FN+6.2-30.38 = 0

FN = -6.2+30.38

FN = 24.18 N

Calculated of the Friction force:

fk=μk*N=  0.2* 24.18 N = 4.836 N

b) We apply the formula (1) to calculated acceleration of the block:

∑Fx = m*ax ,  ax= a  : acceleration of the block

F₁ + F₂x -fk = ( m)*a

65 N + 10.74 -4.836 = ( 3.1)*a

70.904 = ( 3.1)*a

a = (70.904 ) / ( 3.1)

a = 22.87 m/s²

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K.E=1/2mv^2 K.E=1/2multiply1multiply8^2=32joules
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Answer:

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Explanation:

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