In a repair shop a truck engine that has mass 429 kg is held in place by four light cables (Figure 1) . Cable A is horizontal, cables B and D are vertical, and cable C makes an angle of 37.1∘ with a vertical wall. Tension in cable A is 729 N .

a.What is the tension in cable B?
b.What is the tension in cable C?

a. Well, cable A seems to be doing most of the heavy lifting here, so let's focus on that one. We can use some good old trigonometry to solve this. Since cable A is horizontal, it doesn't have any vertical component, which means the sum of the vertical components of the other cables must balance out the weight of the engine (429 kg * g, where g is the acceleration due to gravity). So, let's find the vertical component of cable B!

The tension in cable B can be represented by the equation:

Tension in cable B * sin(90°) = weight of the engine

Wait a minute... sin(90°) is equal to 1! So that means the tension in cable B should be equal to the weight of the engine, which is 429 kg * g. Pretty straightforward, right?

b. Now let's turn our attention to cable C, which is making a funky angle (37.1°) with the vertical wall. The tension in cable C can be represented by the equation:

Tension in cable C * sin(37.1°) = weight of the engine (429 kg * g)

Once again, sin(37.1°) gives us a numerical value, so we can rearrange the equation to solve for the tension in cable C.

Tension in cable C = (429 kg * g) / sin(37.1°)

Voila! You've got your answer! Now, remember, always be careful with those cables. We don't want any clownish accidents in the repair shop!

To solve this problem, we can use Newton's second law, which states that the sum of the forces acting on an object is equal to the mass of the object multiplied by its acceleration. In this case, the object is the truck engine, and the forces acting on it are the tensions in the cables.

Let's break down the given information:
Mass of the truck engine (m) = 429 kg
Tension in cable A (T_A) = 729 N
Angle cable C makes with the vertical wall (θ) = 37.1°

a. Tension in cable B (T_B):
Since cable B is vertical, it only supports the weight of the truck engine. Let's assume the acceleration of the truck engine is zero (since it is held in place). Using Newton's second law, we can write:
Sum of forces in the vertical direction = T_B - m * g = 0
Where g is the acceleration due to gravity (9.8 m/s^2).
Rearranging the equation, we get:
T_B = m * g

Substituting the given values, we get:
T_B = 429 kg * 9.8 m/s^2 ≈ 4204.2 N

Therefore, the tension in cable B is approximately 4204.2 N.

b. Tension in cable C (T_C):
Cable C makes an angle of 37.1° with the vertical wall. We need to resolve the tension in cable C into its vertical and horizontal components.

The vertical component of T_C can be found using the sine function:
Vertical component of T_C = T_C * sin(θ)

The horizontal component of T_C can be found using the cosine function:
Horizontal component of T_C = T_C * cos(θ)

Since the weight of the truck engine is balanced by the vertical component of T_C, we can write:
Vertical component of T_C = m * g

Setting these two equations equal, we get:
T_C * sin(37.1°) = m * g

Substituting the given values and solving for T_C, we get:
T_C = (m * g) / sin(37.1°)

Substituting the known values, we get:
T_C = (429 kg * 9.8 m/s^2) / sin(37.1°) ≈ 6729.2 N

Therefore, the tension in cable C is approximately 6729.2 N.

To find the tension in cable B and cable C, we can apply the equations of equilibrium for the truck engine. These equations will help us understand the forces acting on the engine and cables, leading us to the desired tensions.

a. To find the tension in cable B, we need to consider the vertical forces acting on the truck engine. We start by summing up the forces in the vertical direction:

ΣF_y = 0

The forces acting in the vertical direction are the tension in cable B (T_B), the tension in cable D (T_D), and the weight of the engine (mg), where m is the mass of the engine (429 kg) and g is the acceleration due to gravity (9.8 m/s^2). However, since cable D is perpendicular to the vertical direction, it will not contribute to the vertical forces.

So, our equation becomes:

T_B + mg = 0

Since the weight acts downward, we can rewrite it as:

T_B = -mg

Substituting the known values, we get:

T_B = -(429 kg)(9.8 m/s^2)

Calculating this, we find:

T_B ≈ -4200.6 N

The negative sign indicates that the tension in cable B is acting in the opposite direction to the weight of the engine (upward in this case).

b. To find the tension in cable C, we need to consider the horizontal forces acting on the truck engine. Once again, we start by summing up the forces in the horizontal direction:

ΣF_x = 0

The forces acting in the horizontal direction are the tension in cable A (T_A), the tension in cable C (T_C), and the component of the weight of the engine parallel to the wall. The weight of the engine can be broken down into vertical and horizontal components. The vertical component cancels out due to the equilibrium in the vertical direction, so only the horizontal component remains.

The horizontal component of the weight of the engine can be calculated as:

F_h = mg sin(37.1∘)

So, our equation becomes:

T_A - T_C = F_h

Substituting the known values, we get:

729 N - T_C = (429 kg)(9.8 m/s^2) sin(37.1∘)

Rearranging the equation, we find:

T_C = 729 N - (429 kg)(9.8 m/s^2) sin(37.1∘)

Calculating this, we find:

T_C ≈ 702.4 N

For Part A

T_B = mg + costheta*T_C

T_B = 429*9.8 + cos(37.1)*1250

T_B = 5200N

For part B.

T_A = T_C*(sind(37.1))

729N = T_C*(sind(37.1))

T_C = 1250N