Question

001 (part 1 of 2) 10.0 points
the arm of a crane at a construction site is 17.0 m long, and it makes an angle of 13.4° with the horizontal. assume that the maximum load the crane can handle is limited by the amount of torque the load produces around the base of the arm.
what maximum torque can the crane withstand if the maximum load the crane can handle is 787 n?
answer in units of n·m.

002 (part 2 of 2) 10.0 points
what is the maximum load for this crane at an angle of 38.2° with the horizontal?
answer in units of n.

003 10.0 points

Explanation:

Step1: Recall Torque Formula

Torque \(\tau\) is given by \(\tau = rF\sin\theta\), where \(r\) is the distance (length of crane arm), \(F\) is the force (load), and \(\theta\) is the angle between \(r\) and \(F\). Here, the load is vertical, so \(\theta\) is the angle between the crane arm (r) and horizontal, so the angle between r and vertical (force direction) is \(90^\circ - 13.4^\circ\), but actually, since force is vertical, \(\sin\theta\) where \(\theta\) is the angle between r and horizontal, the vertical component's sine is \(\sin(13.4^\circ)\) (wait, no: torque is \(rF\sin\phi\), where \(\phi\) is the angle between r and the force. The force is weight, vertical, so the angle between r (crane arm, at \(13.4^\circ\) to horizontal) and vertical force is \(90^\circ - 13.4^\circ = 76.6^\circ\)? Wait, no, maybe easier: the perpendicular distance from the base to the line of action of the force is \(r\sin\theta\), where \(\theta\) is the angle between r and horizontal. Because the force is vertical, the horizontal distance is \(r\cos\theta\), vertical distance is \(r\sin\theta\). Wait, torque is \(rF\sin\theta\) where \(\theta\) is the angle between r and the force. The force is vertical, r is at \(13.4^\circ\) to horizontal, so the angle between r and vertical (force) is \(90^\circ - 13.4^\circ = 76.6^\circ\), but \(\sin(76.6^\circ)=\cos(13.4^\circ)\)? No, wait, no: let's draw a right triangle. The crane arm is r, horizontal is x-axis, force is y-axis (downward). The angle between r and x-axis is \(13.4^\circ\), so the angle between r and y-axis (force) is \(90^\circ - 13.4^\circ = 76.6^\circ\). But \(\sin(76.6^\circ)=\cos(13.4^\circ)\)? Wait, no, \(\sin(90^\circ - \alpha)=\cos\alpha\). So \(\sin(76.6^\circ)=\cos(13.4^\circ)\). But wait, torque is also equal to the lever arm (perpendicular distance from axis to force line) times force. The lever arm here is \(r\sin(13.4^\circ)\)? Wait, no: if the crane arm is at angle \(\theta\) to horizontal, the perpendicular distance from the base (axis) to the vertical force (load) is \(r\cos\theta\)? Wait, no, I'm confused. Let's re-express:

Torque \(\tau = rF\sin\theta\), where \(\theta\) is the angle between the position vector (r) and the force vector (F). The position vector is along the crane arm (length r, at \(13.4^\circ\) to horizontal). The force vector is vertical (downward, weight). So the angle between r and F is \(90^\circ - 13.4^\circ = 76.6^\circ\). But \(\sin(76.6^\circ) = \cos(13.4^\circ)\)? Wait, no, \(\sin(90^\circ - \alpha) = \cos\alpha\), so \(\sin(76.6^\circ) = \cos(13.4^\circ)\). But alternatively, the lever arm (perpendicular distance from axis to force) is \(r\sin(13.4^\circ)\)? Wait, no, let's take coordinates: base at origin, crane arm along r at angle \(13.4^\circ\) to x-axis, length r = 17.0 m. The force is at the end of the crane, vertical force F = 787 N downward (y-direction). The torque about the origin is the cross product of r and F. The components of r are \(r\cos\theta\) (x) and \(r\sin\theta\) (y). The force F is (0, -F, 0) (assuming 2D). The torque \(\tau = x(-F) - y(0) = -r\cos\theta F\), but magnitude is \(rF\cos\theta\)? Wait, that can't be. Wait, no, torque is \(rF\sin\phi\), where \(\phi\) is the angle between r and F. r is at \(\theta = 13.4^\circ\) to x-axis, F is at \(270^\circ\) (downward y-axis). The angle between r (13.4° from x-axis) and F (270°) is \(270^\circ - 13.4^\circ = 256.6^\circ\), but sine of that is \(\sin(256.6^\circ) = \sin(180^\circ + 76.6^\circ) = -\sin(76.6^\circ)\), but magnitude is \(rF\sin(76.6^\circ)\). But \(\sin(76.6^\circ) = \cos(13.4^\circ)\)? Wait, no, \(\sin(76.6^\circ) \approx \cos(13.4^\circ)\) because \(76.6 + 13.4 = 90\). So \(\sin(76.6^\circ) = \cos(13.4^\circ)\). Wait, maybe I made a mistake. Let's think differently: the torque is the force times the perpendicular distance from the axis to the line of action of the force. The line of action of the force is vertical (through the end of the crane arm). The perpendicular distance from the base (axis) to this vertical line is the horizontal distance from the base to the end of the crane arm, which is \(r\cos\theta\)? No, wait, the end of the crane arm is at (r cos θ, r sin θ). The vertical line through that point is x = r cos θ. The distance from the origin (axis) to this line is |r cos θ - 0| = r cos θ. But torque is force times this distance, so \(\tau = F \times (r \cos θ)\)? But that would be if the force is horizontal. Wait, no, force is vertical, so the perpendicular distance from axis to vertical line is horizontal distance, which is r cos θ? No, no: for a vertical force, the torque is force times horizontal distance from axis to the point where force is applied. The horizontal distance is r cos θ (since the end of the crane is at x = r cos θ, y = r sin θ). So torque \(\tau = F \times (r \cos θ)\)? But that contradicts the earlier formula. Wait, no, let's use the cross product formula. In 2D, torque is \(x F_y - y F_x\). Here, F is (0, -F), so F_x = 0, F_y = -F. The position vector r is (r cos θ, r sin θ). So torque \(\tau = (r cos θ)(-F) - (r sin θ)(0) = -r F cos θ\). The magnitude is \(r F cos θ\). Wait, but earlier I thought it was \(r F sin θ\). There's a confusion here. Let's clarify: the angle between r and F. r is (r cos θ, r sin θ), F is (0, -F). The angle between r and F is the angle between (cos θ, sin θ) and (0, -1). The dot product is (cos θ)(0) + (sin θ)(-1) = -sin θ. The magnitude of dot product is |r||F|cos φ = r F cos φ, where φ is the angle between them. So -sin θ = cos φ. Therefore, cos φ = -sin θ = cos(90° + θ). So φ = 90° + θ. Then sin φ = sin(90° + θ) = cos θ. Therefore, torque magnitude is \(r F sin φ = r F cos θ\). Ah, so that's correct. So torque is \(r F cos θ\) when θ is the angle between r and horizontal, and force is vertical. Wait, but let's check with θ = 90° (crane arm vertical). Then torque should be zero (since force is along r, so no torque). If θ = 90°, then cos θ = 0, so torque is zero. Correct. If θ = 0° (crane arm horizontal), then torque is \(r F cos 0° = r F\), which is correct (force is vertical, distance is r, so torque is rF). So that formula is correct. So torque \(\tau = r F cos θ\), where θ is the angle between crane arm and horizontal, F is vertical force.

Step2: Plug in Values for Part 1

Given \(r = 17.0\) m, \(F = 787\) N, \(\theta = 13.4^\circ\). So \(\tau = 17.0 \times 787 \times \cos(13.4^\circ)\). First, calculate \(\cos(13.4^\circ)\). Using calculator, \(\cos(13.4^\circ) \approx \cos(13.4) \approx 0.973\) (wait, let's calculate more accurately: 13.4 degrees. Cos(13) is about 0.9744, cos(13.4) is cos(13 + 0.4) = cos(13°)cos(0.4°) - sin(13°)sin(0.4°). But easier to use calculator: 13.4° in radians is 13.4 * π/180 ≈ 0.2338 radians. Cos(0.2338) ≈ 1 - 0.2338²/2 + 0.2338⁴/24 ≈ 1 - 0.0274 + 0.0002 ≈ 0.9728. So approximately 0.9728.

Then \(\tau = 17.0 \times 787 \times 0.9728\). First, 17 * 787 = 13379. Then 13379 * 0.9728 ≈ 13379 * 0.97 ≈ 13379 - 13379*0.03 = 13379 - 401.37 = 12977.63, and 13379 * 0.0028 ≈ 37.46, so total ≈ 12977.63 + 37.46 ≈ 13015.09 N·m. Wait, but let's do it more accurately: 17 * 787 = 13379. 13379 * 0.9728: 13379 * 0.9 = 12041.1, 13379 * 0.07 = 936.53, 13379 * 0.0028 = 37.4612. Sum: 12041.1 + 936.53 = 12977.63 + 37.4612 = 13015.0912 N·m. So approximately 13000 N·m, but let's check the formula again. Wait, maybe I had the angle wrong. Wait, the torque formula is \(\tau = r F \sin\theta\), where θ is the angle between r and the perpendicular to the force. Wait, no, the correct formula is \(\tau = r F \sin\phi\), where \(\phi\) is the angle between r and F. If F is vertical, and r is at angle θ to horizontal, then the angle between r and F is 90° - θ, so \(\sin\phi = \sin(90° - θ) = \cosθ\). So that's consistent. So the formula is correct.

Step3: Solve Part 2

For part 2, we know the maximum torque the crane can withstand is the same as in part 1 (since torque limit is constant). Let's denote \(\tau_{max}\) from part 1. Now, at angle \(\theta' = 38.2^\circ\), we need to find F' such that \(\tau_{max} = r F' \cos\theta'\). So \(F' = \frac{\tau_{max}}{r \cos\theta'}\). First, we need to find \(\tau_{max}\) from part 1. Wait, in part 1, we calculated \(\tau_{max} = 17.0 \times 787 \times \cos(13.4^\circ)\). Let's compute it more accurately. \(\cos(13.4^\circ)\): using calculator, 13.4°: cos(13.4) ≈ 0.972756. So 17 * 787 = 13379. 13379 * 0.972756 ≈ 13379 * 0.972756. Let's compute 13379 * 0.972756:

13379 * 0.9 = 12041.1

13379 * 0.07 = 936.53

13379 * 0.002756 = 13379 * 0.002 + 13379 * 0.000756 = 26.758 + 10.114524 = 36.872524

Sum: 12041.1 + 936.53 = 12977.63 + 36.872524 = 13014.502524 N·m. So \(\tau_{max} ≈ 13014.5\) N·m.

Now, for part 2, \(\theta' = 38.2^\circ\), so \(\cos(38.2^\circ)\): calculate cos(38.2°). 38.2° in radians is 38.2 * π/180 ≈ 0.6667 radians. Cos(0.6667) ≈ 0.7857 (using calculator: cos(38.2°) ≈ 0.7857).

So \(F' = \frac{13014.5}{17.0 \times \cos(38.2^\circ)} = \frac{13014.5}{17.0 \times 0.7857}\). First, 17 * 0.7857 ≈ 13.3569. Then 13014.5 / 13.3569 ≈ 974.3 N. Let's check with more accurate cos(38.2°): using calculator, cos(38.2°) ≈ 0.785678. So 17 * 0.785678 ≈ 13.356526. Then 13014.5 / 13.356526 ≈ 13014.5 ÷ 13.3565 ≈ 974.4 N.

Answer (Part 1):

\(\boxed{13000}\) (approximate, more accurately ~13015 N·m)

Answer (Part 2):

\(\boxed{974}\) (approximate, more accurately ~974 N)

Answer:

(Part 1):
\(\boxed{13000}\) (approximate, more accurately ~13015 N·m)