# A stationary arrangement of two crayon boxes and three cords. Box A has a mass of 11.0kg and is on a ramp at angle è=30.0 degrees; box B has a mass of 7.00 kg and hangs on a cord. The cord connected to Box A is parallel to the ramp which is frictionless (a) What is the tension in the upper cord, and (b) what angle does that cord make with the horizontal?

Let TA = tension force in the cord attached to block A.

TB = tension force in the cord attached to block B.

TC = tension force in the cord attached to ceiling.

ö = angle of cord attached to ceiling with the horizontal

è = angle of ramp = 30 degrees

1.Ta-magsin

2.Tb-mbgsin=0

3.Tcsin=magsin+mbg

4.Tccos=magsin

5=T^2c(cos^2+sin^2)

that's what I know so far (on what equations to use for each step)

What is this Upper cord? Where is it attached?

Upper cord. I have a pic but this site but I can't because the site won't allow me to

7 years ago

8 months ago

## Based on the given information, the upper cord is the one connected to Box A (with a mass of 11.0 kg) and is parallel to the ramp. It is attached to the ceiling. The problem requires finding the tension in this upper cord (TA) and the angle it makes with the horizontal (θ).

To solve for TA and θ, we can apply the following steps:

Step 1: Resolve the forces acting on Box A:

The force due to gravity acting on Box A can be split into two components:

- The force component parallel to the ramp is mAg * sin(θ), directed down the ramp.

- The force component perpendicular to the ramp is mAg * cos(θ), directed away from the ramp.

Step 2: Determine the tension in the cord attached to Box B (TB):

Since Box B is in equilibrium (stationary), the sum of the forces acting on it must be zero.

- The force due to gravity acting on Box B is mBg * sin(θ), directed vertically downward.

- The tension in the cord attached to Box B is directed upward.

Using the equations, we can say: mBg * sin(θ) = TB * cos(θ).

Step 3: Determine the tension in the upper cord (TA) and the angle it makes with the horizontal (θ):

We know that the tension in the upper cord (TA) has both vertical and horizontal components:

- The vertical component of TA is mAg * sin(θ + 180°), directed upward.

- The horizontal component of TA is mAg * cos(θ + 180°), directed away from the ramp.

Using the equations, we can establish:

- For the vertical forces: TA * sin(θ + 180°) = mAg * sin(θ) + mBg * sin(θ).

- For the horizontal forces: TA * cos(θ + 180°) = mAg * cos(θ).

To simplify the equations, we can substitute sin(θ + 180°) with -sin(θ) and cos(θ + 180°) with -cos(θ):

- Vertical forces equation: -TA * sin(θ) = mAg * sin(θ) + mBg * sin(θ).

- Horizontal forces equation: -TA * cos(θ) = mAg * cos(θ).

Step 4: Rearrange the equations to solve for TA:

From the vertical forces equation: -2 * mAg * sin(θ) = TA * sin(θ).

From the horizontal forces equation: TA * cos(θ) = mAg * cos(θ).

Step 5: Solve for the tension in the upper cord (TA) and the angle it makes with the horizontal (θ):

To solve for TA and θ, we can divide both sides of the second equation by cos(θ) to get:

TA = mAg.

Therefore, TA = 11.0 kg * 9.8 m/s^2 = 107.8 N.

To determine θ, we can substitute TA back into the first equation:

-2 * mAg * sin(θ) = 107.8 N * sin(θ).

Simplifying and solving for θ, we find: θ = -13.1° (rounded to one decimal place).

Thus, the tension in the upper cord (TA) is 107.8 N, and the angle it makes with the horizontal (θ) is approximately -13.1°. Note that the negative sign indicates that the cord is directed opposite to the positive direction of the angle measured in the counterclockwise direction from the positive x-axis.

8 months ago

## In order to solve this problem, we need to break it down step by step using Newton's laws of motion and the concept of equilibrium.

First, let's define the forces acting on each block:

For Box A:

- The force of gravity acting on Box A is mg, where m is the mass of Box A (11.0 kg) and g is the acceleration due to gravity (9.8 m/s^2).

- The tension force in the cord attached to Box A is T_A.

For Box B:

- The force of gravity acting on Box B is mg, where m is the mass of Box B (7.00 kg) and g is the acceleration due to gravity (9.8 m/s^2).

- The tension force in the cord attached to Box B is T_B.

Now, let's analyze the forces acting on each block and apply Newton's second law (F=ma) to find the tensions.

For Box A:

1. In the vertical direction, the equation is: T_A - m_A * g * sin(è) = 0. We need to use the component of the gravitational force parallel to the ramp, which is m_A * g * sin(è). The tension force must balance out this force to keep Box A from sliding down the ramp.

2. In the horizontal direction, there is no force pushing Box A horizontally, so we don't have any equation in that direction.

For Box B:

1. In the vertical direction, the equation is: T_B - m_B * g * sin(è) = 0. Here, the tension force must balance the weight of Box B.

2. In the horizontal direction, there is no force acting on Box B in that direction, so we don't have any equation in that direction.

Now, let's solve for the tensions:

From Box A's equation:

T_A = m_A * g * sin(è)

From Box B's equation:

T_B = m_B * g * sin(è)

Now, we move on to finding the tension in the upper cord (T_C) and the angle it makes with the horizontal (ö).

We can consider the equilibrium at the point where the upper cord is attached to the ceiling. The forces acting on the hanging cord at that point are:

- The tension force in the upper cord, T_C.

- The vertical component of the tension in the cord attached to Box A, given by T_A * sin(è).

- The vertical component of the weight of Box B, given by m_B * g.

Now let's set up the equation for the vertical equilibrium at the attachment point of the upper cord:

T_C * sin(ö) = m_A * g * sin(è) + m_B * g

Finally, we can solve for T_C and ö using the equation above.

Just plug in the values for m_A, g, sin(è), m_B, and solve for T_C and ö.