Let <em>F₁ </em>and <em>F₂</em> denote the two forces, and <em>R</em> the resultant force.
<em>F₁ </em>and <em>F₂</em> point perpendicularly to one another, so their dot product is
<em>F₁ </em>• <em>F₂</em> = 0
<em />
We're given that one of these vectors, say <em>F₁</em>, makes an angle with <em>R</em> of 30°, so that
<em>F₁</em> • <em>R</em> = ||<em>F₁</em>|| ||<em>R</em>|| cos(30°)
But we also have
<em>F₁</em> • <em>R</em> = <em>F₁ </em>• (<em>F₁ </em>+ <em>F₂</em>) = (<em>F₁ </em>• <em>F₁</em>) + (<em>F₁ </em>• <em>F₂</em>) = <em>F₁ </em>• <em>F₁ </em>=<em> </em>||<em>F₁</em>||²
So, knowing that ||<em>R</em>|| = 100 N, we get that
(100 N) ||<em>F₁</em>|| cos(30°) = ||<em>F₁</em>||²
(100 N) cos(30°) = ||<em>F₁</em>||
||<em>F₁</em>|| ≈ 86.6 N
(And the same would be true for <em>F₂</em>.)
Answer:
1850 N
Explanation:
The formula for friction force between the load and plane is given as ;
F= μ*N
N = mg cos θ
To find θ, which is the angle the inclined plane makes with the ground at the height of 1.5 m
Sin θ = 1.5/4.5
Sin θ = 0.3333
Sin⁻{0.3333} = 19.50°
θ = 19.50°
Finding N , where m= 500 N , and g= 9.81
N = mg cos θ
N= 500 * 9.81 * cos 19.50°
N= 4624 N
Coefficient of kinetic friction is calculated as;
μ=F/W
μ = 200/500 = 0.4
The magnitude of kinetic friction is given as;
Fk= μ * N
Fk = 0.4 * 4624
Fk= 1850 N
Answer:
Ffriction = 217.7[N]
Explanation:
First, we must find the mass of the box, we must remember that the weight is defined as the product of the mass by acceleration.
where:
w = weight [N] (units of Newtons)
m = mass [kg]
g = gravity acceleration = 9.81 [m/s²]
![1300=m*9.81\\m=132.51[kg]](https://tex.z-dn.net/?f=1300%3Dm%2A9.81%5C%5Cm%3D132.51%5Bkg%5D)
Now we must use Newton's second law, which tells us that the sum of forces is equal to the product of mass by acceleration.
∑F = m*a
where:
∑F sum of forces
m = mass = 132.51[kg]
a = acceleration = 1.3[m/s²]
We must know that the forces that act are the horizontal force that moves the box and the friction force in the negative direction that acts against the movement.
Now replacing:
![390-f_{friction}=132.51*1.3\\390-172.26=f_{friction}\\f_{friction}=217.7[N]](https://tex.z-dn.net/?f=390-f_%7Bfriction%7D%3D132.51%2A1.3%5C%5C390-172.26%3Df_%7Bfriction%7D%5C%5Cf_%7Bfriction%7D%3D217.7%5BN%5D)
Answer:
1768 meters
Explanation:
Distance = speed x time.
(340 m/s) x (5.2 s) = 1768 m
