Question

A 3.00-m-long, 270 N, uniform rod at the zoo is held in a horizontal position by two ropes at its ends in (Figure 1). The left rope makes an angle of 150

∘
with the rod and the right rope makes an angle θ with the horizontal. A 90 N howler monkey (Alouatta seniculus) hangs motionless 0.50 m from the right end of the rod as he carefully studies you. Figure 1 of 1

Answer 1

To solve this problem, we need to analyze the forces acting on the rod and apply the principles of equilibrium.

Step-by-step explanation:

The student is asking about a situation where a 3.00-m-long, 270 N, uniform rod is held in a horizontal position by two ropes. One rope is at a 150° angle with the rod and the other rope is at an angle θ with the horizontal. A 90 N howler monkey hangs motionless 0.50 m from the right end of the rod. To solve this problem, we need to analyze the forces acting on the rod and apply the principles of equilibrium.

Answer 2

The tension in the left rope is approximately 120 N.

Analyze the forces:

There are three forces acting on the rod:

Tension in the left rope (T1)

Tension in the right rope (T2)

Weight of the monkey (W) acting downwards at 0.5 m from the right end.

2. Apply equilibrium conditions:

Since the rod is in static equilibrium (not moving or accelerating), the sum of forces and torques must be zero.

3. Sum of forces in the vertical direction:

T1y + T2y - W = 0

4. Sum of forces in the horizontal direction:

T1x = 0 (as the left rope makes an angle of 150° with the horizontal, its horizontal component is zero)

5. Sum of torques around the pivot point (right end of the rod):

(0.5 m)W - (3.00 m)T1y = 0 (counterclockwise torques are positive)

6. Solve the system of equations:

We have three equations and three unknowns (T1y, T2y, and T2x).

From equation 4, we know T1x = 0.

Substitute this into equation 3: T1y + T2y - W = 0.

From equation 5, we can express T1y as: T1y = (0.5 m * W) / (3.00 m).

Substitute this expression for T1y in the modified equation 3: (0.5 m * W) / (3.00 m) + T2y - W = 0.

Solve for T2y: T2y = (2/3)W.

7. Calculate the tension in the left rope (T1):

Use the trigonometric relationship for the left rope: sin(150°) = T1y / T1.

Since sin(150°) = -0.5, we have -0.5 = (2/3)W / T1.

Solve for T1: T1 = (2/3)W / (-0.5).

Substitute the weight of the monkey (W = 90 N): T1 = (2/3) * 90 N / (-0.5).

Calculate the tension: T1 ≈ 120 N.