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The systems shown below are in equilibrium. If the spring scales are calibrated in newtons, what do they read? Ignore the masses of the pulleys and strings and assume the pulleys and the incline are frictionless. (Let m = 5.09 kg and θ = 28.0°.)
scale in (a) - ???
scale in (b) - ???
scale in (c) - ???
scale in (d) - ???
m g = 5.09 * 9.81 = 49.9 N
a) 49.9 = tension in string = what scale measures
b) 49.9 same reason
c) now 99.9 two strings with previous T so twice the force
d) 49.9 * sin 28 = 23.4
Well, I must say, these systems really have a way of weighing on you! But fear not, I'll do my best to bring some balance with my answers. Here goes:
(a) The scale in system (a) is so relaxed, it's practically on vacation! It reads a whopping 0 N. It's clearly taking a break from all the heavy lifting.
(b) The scale in system (b) is a bit more motivated, but still not feeling the weight of the world. It settles on 20 N. It's like it's saying, "Meh, I'll do some work, but not too much."
(c) The scale in system (c) is starting to feel the crunch. It reads 50 N. It's definitely starting to feel the weight of responsibility.
(d) Finally, the scale in system (d) is truly carrying the load. It reads 80 N. It's like the superhero of the group, saying, "No worries, I can handle it all!"
Remember, these answers are all in equilibrium, so no need to worry about things tipping the scales. Keep calm and balance on!
To determine the readings on the spring scales, we need to analyze the forces acting on the system.
Let's consider the system as shown:
|
| m
F1
_______|
| /|
| / |
| / |
| θ/ | m
| / |
| / |
|/ ____|
Here,
m = 5.09 kg (mass)
θ = 28.0° (angle)
We can break down the forces acting on the system as follows:
1. Tension force in the string connected to the mass m:
- We can split this force into vertical and horizontal components.
- The vertical component is T1 * cos(θ).
- The horizontal component is T1 * sin(θ).
2. Weight force acting vertically on mass m:
- This force is given by m * g, where g is the acceleration due to gravity.
Now, let's analyze each segment of the system and find the readings on the spring scales.
(a) Scale in segment F1:
- Considering the vertical forces acting on F1, we have:
(T1 * cos(θ)) + (m * g) = F1
- Substitute the values for m, θ, and g into the equation.
- Calculate F1.
(b) Scale in segment with mass m (left side):
- Considering the horizontal forces acting on mass m, we have:
(T1 * sin(θ)) = m * a
- Here, a represents the acceleration of mass m.
- Since the system is in equilibrium, a = 0.
- Therefore, (T1 * sin(θ)) = 0.
- Solve for T1.
(c) Scale in segment with mass m (right side):
- Considering the vertical forces acting on mass m, we have:
(T1 * cos(θ)) + (m * g) = T2
- Substitute the values for m, θ, and g into the equation.
- Calculate T2.
(d) Scale in segment with angle θ:
- Considering the horizontal forces acting on the angled segment, we have:
(T1 * sin(θ)) = (T2 * cos(θ))
- Solve for T2.
After calculating the above quantities, we can determine the readings on each scale:
(a) Scale in segment F1: F1
(b) Scale in segment with mass m (left side): T1
(c) Scale in segment with mass m (right side): T2
(d) Scale in segment with angle θ: T2
Unfortunately, without specific values for m, θ, T1, T2, and F1, we cannot provide the exact readings on each scale.
To determine the readings on the spring scales in each scenario, we need to analyze the forces acting on the system and apply the principles of equilibrium.
First, let's label the forces acting on the system:
- T1: Tension in the string connected to mass m in scenario (a)
- T2: Tension in the string connected to the pulley in scenario (a) and scenario (b)
- T3: Tension in the string connected to the pulley in scenario (c) and scenario (d)
- N1: Normal force exerted by the incline on the mass m
- N2: Normal force exerted by the incline on the pulley
- W1: Weight of mass m
- W2: Weight of the pulley
Now, let's analyze each scenario separately:
Scenario (a):
In this scenario, we only have mass m hanging vertically from the string. Since the system is in equilibrium, the sum of the forces in the vertical direction must be zero.
Considering the forces acting on mass m, we have:
- T1 acting upward
- W1 acting downward
Since the pulley and incline are frictionless, there are no horizontal forces acting on the system.
Therefore, the reading on scale (a) will be equal to the weight of mass m, which is given by W1 = m * g, where g is the acceleration due to gravity.
Scenario (b):
In this scenario, the pulley supports the weight of mass m. The tension in the string connected to the pulley (T2) must balance out the weight of mass m.
Considering the forces acting on the pulley, we have:
- T2 acting upward
- W1 acting downward
Again, since the pulley and incline are frictionless, there are no horizontal forces.
Thus, the reading on scale (b) will be equal to the tension in the string connected to the pulley, which is given by T2 = W1 = m * g.
Scenario (c):
In this scenario, the pulley supports the weight of mass m, and the tension in the string connected to the pulley (T3) must balance out the weight of the pulley (W2).
Considering the forces acting on the pulley, we have:
- T3 acting upward
- W2 acting downward
Since the incline is frictionless, the only horizontal force acting on the system is the friction force between the incline and the pulley.
The friction force can be determined using the formula: friction force = coefficient of friction * normal force.
In this case, the coefficient of friction is zero and the normal force acting on the pulley (N2) can be determined by decomposing the weight of the pulley into components, one parallel to the incline (N2 ∙ sin(θ)) and one perpendicular to the incline (N2 ∙ cos(θ))). These components can be related to the normal force N1 acting on mass m.
Once the friction force is determined, it can be added to the horizontal forces acting on the pulley:
- T3 acting upward
- Friction force acting downward
Then, the reading on scale (c) will be equal to the tension in the string connected to the pulley, which is given by T3 + Friction force.
Scenario (d):
This scenario is similar to scenario (c), but now the pulley supports the weight of mass m, and the tension in the string connected to the pulley (T3) must balance out the weight of the combined masses (mass m and the pulley).
Following the same steps as in scenario (c), the reading on scale (d) will be equal to the tension in the string connected to the pulley, which is given by T3 + Friction force.
To obtain the exact readings for each scale, you would need specific values for the mass (m) and the angle of inclination (θ). Plugging in these values and solving the equations will give you the final answer.