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A crane has an arm length of 20m inclined at 30deg with the vertical. It carries a container of mass of 1 ton suspended from the top end of the arm. Find the torque produced by the gravitational force on the container about the point where the arm is fixed to the crane. Given: 1 ton = 1000 kg; neglect the weight of the arm. Hint: take the force and perpendicular distance from the point where the arm is fixed to the crane. Note: Express your answer in whole number. No unit is required for the final answer. Set your calculator in radians.

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To find the torque produced by the gravitational force on the container, we need to calculate the force and perpendicular distance from the point where the arm is fixed to the crane.

First, let's calculate the force exerted by the gravitational force on the container. The weight of the container can be calculated using its mass and the acceleration due to gravity. Given that the mass of the container is 1000 kg (1 ton) and the acceleration due to gravity is approximately 9.8 m/s^2, we can calculate the force as follows:

Force = mass * acceleration due to gravity
Force = 1000 kg * 9.8 m/s^2
Force = 9800 N

Next, we need to find the perpendicular distance from the point where the arm is fixed to the crane to the line of action of the gravitational force. This distance can be determined using trigonometry.

The perpendicular distance can be calculated using the formula:

Perpendicular distance = arm length * sin(angle)

Given that the arm length is 20 m and the angle with the vertical is 30 degrees, we can calculate the perpendicular distance as follows:

Perpendicular distance = 20 m * sin(30 degrees)
Perpendicular distance = 20 m * 0.5
Perpendicular distance = 10 m

Now, we can calculate the torque using the formula:

Torque = force * perpendicular distance

Substituting the values we calculated earlier:

Torque = 9800 N * 10 m
Torque = 98000 Nm

Therefore, the torque produced by the gravitational force on the container about the point where the arm is fixed to the crane is 98000 Nm.
User Ranju
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