Question

A load of bricks of mass \(M=200.0 \mathrm{~kg}\) is attached to a crane by a cable of negligible mass and length \(L=3.00 \mathrm{~m}\). Initially, when the cable hangs vertically downward, the bricks are a horizontal distance \(D=1.50 \mathrm{~m}\) from the wall where the bricks are to be placed. What is the magnitude of the horizontal force that must be applied to the load of bricks (without moving the crane) so that the bricks will rest directly above the wall?

Short answer

Answer: The required horizontal force is approximately 981 N.

Step-by-step solution

Determine the cable tension

We need to determine the tension in the cable. When the crane suspends the load of bricks, it is in static equilibrium. To achieve this equilibrium, the upward force exerted by the cable (cable tension) must be equal to the downward force due to the bricks' weight. Therefore, the tension, T, will equal the weight of the bricks, which can be calculated using the equation: T = M * g, where M is the mass of the bricks and g is the acceleration due to gravity (approximately 9.81 m/s^2). T = M * g T = 200 kg * 9.81 m/s^2 T ≈ 1962 N The tension in the cable is approximately 1962 N.

Determine the angle between the cable and the wall

To determine the horizontal force that must be applied to the bricks, we must first find the angle between the cable and the wall. In the initial position, the cable forms a right-angled triangle with the wall, with the horizontal distance D and the cable length L as the legs and the vertical distance as the hypotenuse. We can use the cosine of the angle between the cable and the wall to find this angle: cos(θ) = (D) / (L) cos(θ) = (1.5 m) / (3.0 m) cos(θ) = 0.5 θ ≈ 60° The angle between the cable and the wall is approximately 60 degrees.

Calculate the horizontal component of tension

Now that we have the angle and the cable tension, we can use the horizontal component of the tension to find the required horizontal force. This component can be calculated using the formula: T_horizontal = T * cos(θ) T_horizontal ≈ 1962 N * cos(60°) T_horizontal ≈ 981 N The horizontal component of tension is approximately 981 N.

Determine the horizontal force

Finally, to achieve static equilibrium, the horizontal force that must be applied to the load of bricks should be equal and opposite to the horizontal component of tension, as determined in Step 3. Therefore, the magnitude of the horizontal force needed is: F_horizontal = T_horizontal F_horizontal ≈ 981 N The required horizontal force to make the bricks rest directly above the wall is approximately 981 N.