The horizontal force is m*v²/Lh, where m is the total mass. The vertical force is the total weight (233 + 840)N.
<span>Fx = [(233 + 840)/g]*v²/7.5 </span>
<span>v = 32.3*2*π*7.5/60 m/s = 25.37 m/s </span>
<span>The horizontal component of force from the cables is Th + Ti*sin40º and the vertical component of force from the cable is Ta*cos40º </span>
<span>Thh horizontal and vertical forces must balance each other. First the vertical components: </span>
<span>233 + 840 = Ti*cos40º </span>
<span>solve for Ti. (This is the answer to the part b) </span>
<span>Horizontally </span>
<span>[(233 + 840)/g]*v²/7.5 = Th + Ti*sin40º </span>
<span>Solve for Th </span>
<span>Th = [(233 + 840)/g]*v²/7.5 - Ti*sin40º </span>
<span>using v and Ti computed above.</span>
Answer: Tension = 47.8N, Δx = 11.5×
m.
Tension = 95.6N, Δx = 15.4×
m
Explanation: A speed of wave on a string under a tension force can be calculated as:
is tension force (N)
μ is linear density (kg/m)
Determining velocity:
0.0935 m/s
The displacement a pulse traveled in 1.23ms:
Δx = 11.5×
With tension of 47.8N, a pulse will travel Δx = 11.5× m.
Doubling Tension:
|v| = 0.1252 m/s
Displacement for same time:
15.4×
With doubled tension, it travels 15.4× m
Answer:
Explanation:
Let's look at a mathematical representation of this. The equation for tis is just a souped up version of Newton's 2nd Law:
F - f = ma. It an object is moving at a constant speed, the acceleration of that object is 0. That changes this equation to
F = f which states that the applied Force equals the frictional force, choice a.
Answer:
m = 1 kg
Explanation:
Given that,
The force constant of the spring, k = 39.5 N/m
The frequency of oscillation, f = 1 Hz
The frequency of oscillation is given by the formula as formula as follows :
So, the mass that is attached to the spring is 1 kg.
<span>Answer:
Therefore, x component: Tcos(24°) - f = 0 y component: N + Tsin(24°) - mg = 0 The two equations I get from this are: f = Tcos(24°) N = mg - Tsin(24°) In order for the crate to move, the friction force has to be greater than the normal force multiplied by the static coefficient, so... Tcos(24°) = 0.47 * (mg - Tsin(24°)) From all that I can get the equation I need for the tension, which, after some algebraic manipulation, yields: T = (mg * static coefficient) / (cos(24°) + sin(24°) * static coefficient) Then plugging in the values... T = 283.52.
Reference https://www.physicsforums.com/threads/difficulty-with-force-problems-involving-friction.111768/</span>
