Math  /  Algebra

QuestionA scaffold of mass 64 kg and length 7.9 m is supported in a horizontal position by a vertical cable at each end. A window washer of mass 88 kg stands at a point 2.8 m from one end. What is the tension in (a) the nearer (relative to the person) cable and (b) the farther (relative to the person) cable? (a) Number
Units \square (b) Number \square Units \square

Studdy Solution

STEP 1

What is this asking? A window washer is standing on a scaffold supported by two cables, and we need to find the tension (force) in each cable. Watch out! Don't forget to consider the weight of *both* the scaffold and the window washer, and where their weights are centered!

STEP 2

1. Define key values
2. Calculate the total weight
3. Set up the torque equation
4. Solve for the tension in the farther cable
5. Solve for the tension in the nearer cable

STEP 3

Alright, let's **define** our key values!
We have the mass of the scaffold, ms=64 kgm_s = 64 \text{ kg}, and its length, L=7.9 mL = 7.9 \text{ m}.
The window washer's mass is mw=88 kgm_w = 88 \text{ kg}, and they are standing d=2.8 md = 2.8 \text{ m} from one end.

STEP 4

Now, let's **calculate** the weights!
Remember, weight is mass times the acceleration due to gravity, which we'll approximate as g=9.8 m/s2g = 9.8 \text{ m/s}^2.
The weight of the scaffold is Ws=msg=64 kg9.8 m/s2=627.2 NW_s = m_s \cdot g = 64 \text{ kg} \cdot 9.8 \text{ m/s}^2 = 627.2 \text{ N}.
The weight of the window washer is Ww=mwg=88 kg9.8 m/s2=862.4 NW_w = m_w \cdot g = 88 \text{ kg} \cdot 9.8 \text{ m/s}^2 = 862.4 \text{ N}.

STEP 5

Here's where the magic happens!
We'll use the concept of **torque** to solve this problem.
Torque is a rotational force, and in this case, the total torque around any point on the scaffold must be zero since the scaffold isn't rotating.
Let's choose the point where the *nearer* cable attaches as our pivot point.

STEP 6

The **torque** due to the scaffold's weight acts at its center, which is L/2=7.9 m/2=3.95 mL/2 = 7.9 \text{ m} / 2 = 3.95 \text{ m} from the nearer cable.
The torque due to the window washer's weight acts at a distance of d=2.8 md = 2.8 \text{ m} from the nearer cable.
The tension in the farther cable, TfT_f, acts at a distance of L=7.9 mL = 7.9 \text{ m} from the nearer cable.

STEP 7

The torque equation is the sum of all torques, which must equal zero.
Remember, clockwise torques are negative, and counter-clockwise torques are positive.
So, we have Ws(L/2)Wwd+TfL=0 -W_s \cdot (L/2) - W_w \cdot d + T_f \cdot L = 0.

STEP 8

Let's **plug in** our values: 627.2 N3.95 m862.4 N2.8 m+Tf7.9 m=0-627.2 \text{ N} \cdot 3.95 \text{ m} - 862.4 \text{ N} \cdot 2.8 \text{ m} + T_f \cdot 7.9 \text{ m} = 0.
This simplifies to 2478.64 Nm2414.72 Nm+7.9Tf m=0-2478.64 \text{ Nm} - 2414.72 \text{ Nm} + 7.9 T_f \text{ m} = 0.

STEP 9

Adding the negative terms and dividing both sides by 7.9 m7.9 \text{ m}, we get Tf=(4893.36 Nm)/(7.9 m)=620 NT_f = (4893.36 \text{ Nm}) / (7.9 \text{ m}) = 620 \text{ N}.
So, the tension in the farther cable is 620 N620 \text{ N}.

STEP 10

Now for the **nearer cable**!
The total upward force must equal the total downward force.
So, the tension in the nearer cable, TnT_n, plus the tension in the farther cable, TfT_f, must equal the total weight: Tn+Tf=Ws+WwT_n + T_f = W_s + W_w.

STEP 11

Plugging in our values, we have Tn+620 N=627.2 N+862.4 NT_n + 620 \text{ N} = 627.2 \text{ N} + 862.4 \text{ N}, which simplifies to Tn+620 N=1489.6 NT_n + 620 \text{ N} = 1489.6 \text{ N}.
Subtracting 620 N620 \text{ N} from both sides gives us Tn=870 NT_n = 870 \text{ N}.

STEP 12

The tension in the nearer cable is **870 N**, and the tension in the farther cable is **620 N**.

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