A crate of mass 2000 kg is hanging from a crane at the end of a cable 12 m long. If we attach a horizontal rope to this crate and gradually apply a pull of 1800 N, what angle will the cable finally make with the vertical?
45 degree angle, i think
To find the angle that the cable finally makes with the vertical, we can use trigonometry. Let's break down the forces acting on the crate:
1. The weight acting downwards (W): The weight of the crate is given by the formula W = m * g, where m is the mass of the crate in kg and g is the acceleration due to gravity (approximately 9.8 m/s^2).
W = 2000 kg * 9.8 m/s^2 = 19,600 N
2. The tension in the horizontal rope (T): Initially, T is zero because no force is acting horizontally. As we gradually apply a pull of 1800 N, the tension in the rope will become 1800 N.
Now, let's draw the forces on a diagram:
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W = 19,600 N (downwards)
T = 1800 N (horizontally)
Now, let's consider the forces in the vertical direction (the y-axis). The vertical component of tension (T_y) and the weight (W) must balance each other out because the crate is at rest. This means:
T_y = W
Since the cable makes an angle with the vertical, we can write:
T_y = T * cos(theta)
where theta is the angle between the cable and the vertical.
Now, let's substitute the known values into the equation:
T * cos(theta) = W
1800 N * cos(theta) = 19,600 N
Solving for cos(theta), we have:
cos(theta) = 19,600 N / 1800 N
cos(theta) = 10.89
To find the angle theta, we need to take the inverse cosine (cos^-1) of 10.89:
theta = cos^-1(10.89)
Using a calculator or trigonometric table, we find that cos^-1(10.89) is not a real value. This indicates that the cable cannot make a vertical angle under the given conditions.
Therefore, there is no angle for the cable with the vertical that satisfies the given scenario.