Wanda is working out with weights and manages to lift a...

Wanda is working out with weights and manages to lift a light rope with a $10 \mathrm{~kg}$ mass hanging from it. When she is through lifting the right side of the rope and the left side of the rope each make an angle of $\theta=15^{\circ}$ with respect to the horizontal. See Fig. 6-80. (a) Draw a free-body diagram showing the forces on the midpoint of the rope (where it is the lowest). (b) What are the magnitudes of each of her pulling forces $\vec{F}_{A}$ and $\vec{F}_{B}$ ? (c) How hard would Wanda have to pull with each hand to raise the $10 \mathrm{~kg}$ mass so that the rope becomes perfectly horizontal?

Wanda is working out with weights and manages to lift a light rope with a $10 \mathrm{~kg}$ mass hanging from it. When she is through lifting the right side of the rope and the left side of the rope each make an angle of $\theta=15^{\circ}$ with respect to the horizontal. See Fig. 6-80.
(a) Draw a free-body diagram showing the forces on the midpoint of the rope (where it is the lowest).
(b) What are the magnitudes of each of her pulling forces $\vec{F}_{A}$ and $\vec{F}_{B}$ ?
(c) How hard would Wanda have to pull with each hand to raise the $10 \mathrm{~kg}$ mass so that the rope becomes perfectly horizontal?

Key Concepts

Free-Body Diagram

A free-body diagram is a graphical representation that depicts all the forces acting on a single object, isolated from its environment. It is used to simplify the analysis of an object's motion or equilibrium by showing the magnitude and direction of forces such as gravity, applied forces, normal forces, and friction.

Static Equilibrium

Static equilibrium refers to the situation in which an object is at rest, and the sum of all forces and torques acting on it is zero. This concept is essential in determining the conditions under which an object remains stationary, implying that every force component in each direction must cancel out.

Force Decomposition

Force decomposition involves breaking a force vector into its horizontal and vertical components using trigonometric functions such as sine and cosine. This technique is crucial when analyzing forces acting at an angle, allowing the resolution of complex force interactions into manageable scalar equations.

Tension in Ropes

Tension in ropes refers to the force transmitted through a rope, cable, or string when it is pulled tight by forces acting at its ends. In problems involving ropes, understanding how tension varies with angles and how it supports weights or balances forces is key to analyzing and solving the system's equilibrium.

Transcript

00:01 All right, so we've got this mass that is being pulled up by wanda, and the mass is 10 kilograms, and she pulls with both hands to a point where each rope, each end of the rope, makes an angle of 15 degrees with the horizontal.

00:19 And we're going to start by just drawing a force diagram of this middle point here.

00:25 So the force is acting on that middle part of the string.

00:29 We have fc down.

00:31 Which is the weight force of the mass.

00:34 We're going to have a which is pulling up at that angle of 15 degrees and we've got b which is in the exact opposite direction and not exact opposite and the other side also with 15 degrees to the horizontal so our free body diagram is going to look like that.

00:56 Part b we need to figure out what the values are these and i'm going to start in the x direction because the force the force is acting on in both directions x and y will cancel it will all end up being zero there's no net force acting on the mass because she's able to lift it up and hold it there and so the net forces in the y direction and the x direction are all going to equal zero so the the sum of the forces in the x direction are going to be this applied force a in the left direction, which will believe that as positive because it's a.

01:41 So f, a times the cosine of that angle, 15 degrees.

01:47 And then in the wide, in the other direction, we have force b also cosine of that angle.

01:55 So these are going to end up being equal and opposite to each other.

02:01 F .a.

02:02 Cosine theta has equal fb cosine theta? well, there's a cosine theta.

02:09 These are the same.

02:10 So those cosine thetas, those cancel each other out.

02:13 And we just have that f -a equals f -b.

02:18 Now that's intuitively obvious, but i wanted to work through the math to show that it is indeed the case.

02:27 So the amount of force she's applying with her left hand has to equal the amount of force she's applying with the right hand.

02:34 So that will come in handy when we are evaluating the forces in the y direction.

02:40 So in the y direction, again, these are going to be equal to zero...

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