The systems shown below are in equilibrium (with m = 4.80 kg and θ = 27.0°). 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.

scale in (c)
scale in (d)

http://einstein.drexel.edu/~wking/courses/phys101_w07/hwk/hwk4.pdf

To determine the readings on the spring scales, we need to analyze the forces acting on the system. Let's break down the tensions and forces in each component:

(a) Tension in the vertical string:
Since the masses on the left and right sides are in equilibrium, the tension in the vertical string must balance the weight of both masses:
Tension = m * g
Tension = 4.80 kg * 9.8 m/s^2
Tension ≈ 47.04 N

(b) Tension in the horizontal string:
The horizontal string connects the vertical string to the pulley. Since the pulley is frictionless, the tension in the horizontal string will be the same as the tension in the vertical string:
Tension ≈ 47.04 N

(c) Scale reading on the scale in position (c):
The scale in position (c) is attached to the horizontal string. Since the tension in the horizontal string is approximately 47.04 N, the scale will read:
Scale reading = 47.04 N

(d) Scale reading on the scale in position (d):
The scale in position (d) is attached to the right mass. The force acting on the right mass is the component of its weight along the incline, which is given by:
Force = m * g * sin(θ)
Force = 4.80 kg * 9.8 m/s^2 * sin(27.0°)
Force ≈ 22.54 N

Therefore, the scale in position (d) will read approximately 22.54 N.

To determine the readings on the spring scales in both (c) and (d), we need to analyze the forces acting on each system and apply Newton's laws of motion.

Let's break down the forces acting on each system individually:

System in (c):
In this system, there are two forces acting on the mass m: the gravitational force (mg) acting vertically downward and the tension in the string (T) acting at an angle θ with respect to the horizontal.

To find the tension in the string, we need to resolve the gravitational force into its components. The component perpendicular to the incline is mg * cos(θ) and the component parallel to the incline is mg * sin(θ).

The tension in the string is equal to the net force acting parallel to the incline, which can be calculated using the equation:
mg * sin(θ) - T = 0

Solving this equation for T, we get:
T = mg * sin(θ)

System in (d):
In this system, there are three forces acting on the mass m: the gravitational force (mg) acting vertically downward, the tension in the upper string (T₁), and the tension in the lower string (T₂).

Again, we need to resolve the gravitational force into its components. The component perpendicular to the incline is mg * cos(θ) and the component parallel to the incline is mg * sin(θ).

Now, let's apply Newton's laws of motion to the system in (d). Newton's second law states that the net force acting on an object is equal to its mass times its acceleration. Since the system is in equilibrium, the acceleration is zero, which means the net force is also zero.

Considering the forces acting on the mass m, we can write the following equations:
T₂ - mg * cos(θ) = 0 (since there is no acceleration in the vertical direction)
T₁ - T₂ - mg * sin(θ) = 0 (since there is no acceleration in the horizontal direction)

Solving these equations simultaneously, we can find the values of T₁ and T₂.

Now that we have determined the equations for both systems, we can substitute the given values (m = 4.80 kg and θ = 27.0°) to find the values of T in each case.

Once you calculate the values for T in both systems, those will be the readings on the respective spring scales in newtons.