Two ropes are attached a wagon one horizontal to the west with a tension force of 75N and other east and at an angle of 38 degrees upward from horizontal with a lension force of 125 N Find the components of the net force on the cart Show your work
Answers
Answer:
The components of the net force acting on the cart are;
Horizontal force, Fₓ ≈ 23.5 N acting east
The vertical force, [tex]F_y[/tex] ≈ 76.96 N
Explanation:
The parameters given in the question are;
The magnitude of the force on the rope acting horizontally to the west = 75 N
The magnitude of the force on the other rope east acting at an angle to the horizontal = 125 N
The angle to the horizontal of the force on the other rope east to the horizontal = 38°
Therefore, we have;
The horizontal component of the force acting east = 125 × cos(38°) ≈ 98.5 N
The net sum of the horizontal forces, ∑Fₓ noting that the horizontal forces are acting in opposite direction is given as follows;
∑Fₓ = 75 - 125 × cos(38°) = 75 - 98.3 = -23.5N
Given that the difference in forces is negative, and hat we took the force acting east as negative, we have that the net horizontal force is acting east with a horizontal component of 23.5N
The net vertical force, ∑F[tex]_y[/tex], is given as follows;
∑F[tex]_y[/tex] = 125 × sin(38°) ≈ 76.96 N
Therefore, the components of the net force acting on the cart are;
Horizontal force, Fₓ ≈ 23.5 N, the vertical force, [tex]F_y[/tex] ≈ 76.96 N
The components of the net force on the cart are 23.5 Newton and 76.96 Newton.
Given the following data:
Tension force (horizontal) due west = 75 NTension force (horizontal) due east = 125 NAngle of inclination = 38°To find the components of the net force on the cart:
First of all, we would determine the horizontal component of the force acting due east.
[tex]F_y = Fcos(\theta)\\\\F_y = 125cos(38)\\\\F_y = 125 \times 0.7880\\\\F_y = 98.5 \; Newton[/tex]
Next, we would find the resultant horizontal forces:
[tex]\sum F_y = F + F_y\\\\\sum F_y = 75 + (- 98.5)\\\\\sum F_y = -23.5 \;Newton[/tex]
Therefore, the horizontal component of the net force acts in the opposite direction due east.
Horizontal component of the net force = 23.5 Newton
For the vertical component of the net force:
[tex]\sum F_x = Fsin(\theta)\\\\\sum F_x = 125cos(38)\\\\ \sum F_x = 125 \times 0.6157\\\\ \sum F_x = 76.96\; Newton[/tex]
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Explanation:
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Unknown:
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According to newtons second law of motion;
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Answer:
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Answer:
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Answer:
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Answers
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Answer:
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Answer:
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Answer:
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Answers
Answer:
The answer is 4 m/s²Explanation:
The acceleration of an object given it's mass and the force acting on it can be found by using the formula
[tex]a = \frac{f}{m} \\ [/tex]
where
f is the force
m is the mass
From the question
f = 20 N
m = 5 kg
We have
[tex]a = \frac{20}{5} \\ [/tex]
We have the final answer as
4 m/s²Hope this helps you
Find the mass if the force is 18 N and the acceleration is 2 m/s2.
Answers
Answer:
The answer is 9 kgExplanation:
To find the mass given the force and acceleration we use the formula
[tex]m = \frac{f}{a} \\ [/tex]
where
f is the force
a is the acceleration
We have
f = 18 N
a = 2 m/s²
We have
[tex]m = \frac{18}{2} \\ [/tex]
We have the final answer as
9 kgHope this helps you
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Answers
Answer:
45m/s^1
Explanation:
v=u+at
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Answer:
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Explanation:
Answer:
Newton's first law is also called the law of Inertia.
Explanation:
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Mainly there ae three types of inertia-
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What is Newton's first law of motion ?If a body is in motion or in rest, it wants to remain in its position unless any external force is applied.
Thus Newton's first law of motion is also called law of inertia.
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