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A ladder carried by a fire truck is \(20.0 \mathrm{~m}\) long. The ladder weighs \(3400 \mathrm{~N}\) and its center of gravity is at its center. The ladder is pivoted at one end (A) about a pin (Fig. E11.7); ignore the friction torque at the pin. The ladder is raised into position by a force applied by a hydraulic piston at \(C\). Point \(C\) is \(8.0 \mathrm{~m}\) from \(A,\) and the force \(\overrightarrow{\boldsymbol{F}}\) exerted by the piston makes an angle of \(40^{\circ}\) with the ladder. What magnitude must \(\boldsymbol{F}\) have to just lift the ladder off the support bracket at \(B ?\) Start with a free-body diagram of the ladder.

Short Answer

Expert verified
The required force, F, to just lift the ladder off the support bracket at B is approximately \(7930.6 N\).

Step by step solution

01

Determine Torque due to Weight of the Ladder

This requires use of the law of torques, which states that the sum of the torques on an object is equal to the product of the force and the perpendicular distance to the pivot point (torque = force * distance). The distance from the center of gravity of the ladder (midpoint) to the pivot point A is 10 m. Thus, the torque due to the weight of the ladder is \( \tau_w = F_w * d_w = (3400 N) * (10 m) = 34000 N.m \).
02

Calculate the Horizontal Component of the Force F

The force from the hydraulic piston, F, is applied at an angle of 40 degrees to the ladder. This creates two components of force: horizontal and vertical. The horizontal component (F_h) does not contribute to the torque because it is applied parallel to the ladder, hence will not work on rotating the ladder around point A. Therefore it's important to figure it out and it's given by \( F_h = F * \cos(40^\circ) \).
03

Calculate the Vertical Component of the Force F

The vertical component (F_v) of the force works to lift the ladder around point A. It's calculated as \( F_v = F * \sin(40^\circ) \).
04

Calculate the Total Torque due to the Force F and Set up the Equation

The total torque due to force F is equal to the product of the vertical component of F and the distance from point C to A, which gives \( \tau_F = F_v * d_c = F * \sin(40^\circ) * (8 m) \). To just start lifting the ladder off the support at B - we need to make sure that the torque due to the weight of ladder equals the torque due to the force F. Thereby we have \( \tau_w = \tau_F \). Substituting the calculated values we get \( 34000 N.m = F * \sin(40^\circ) * (8 m) \).
05

Solve for F

Rearrange the equation from step 4 to solve for F. This results in \( F = \frac{34000 N.m}{\sin(40^\circ) * (8 m)} = 7930.6 N \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Torque
Torque is the rotational equivalent of force, crucial for understanding static equilibrium, such as the ladder in our exercise. When a force is applied at a distance from a pivot point, it creates torque, which can cause an object to rotate. The amount of torque depends on two main factors: the magnitude of the force and the distance from the pivot, known as the lever arm.

In our ladder exercise, torque comes into play due to the weight of the ladder and the force exerted by the hydraulic piston. The torque due to the ladder's weight acts at its center of gravity, halfway along its length (10 m from the pivot), and is calculated as the product of the force (weight of the ladder) and the distance to the pivot:
  • Weight of ladder: 3400 N
  • Distance (lever arm): 10 m
  • Torque: 34000 N·m
This calculated torque helps us determine the required force by the piston necessary to lift the ladder.
Force Components
Forces can act in different directions and often must be broken down into components to analyze their effects fully. Understanding force components is crucial when forces do not line up perfectly with the axes of motion, as with the hydraulic piston force applied at an angle in the exercise.

The force from the piston can be split into two components: horizontal and vertical. These components are found using trigonometric functions:
  • Horizontal component (F_h) is parallel to the ladder and given by: \( F_h = F \cdot \cos(40^\circ) \)
  • Vertical component (F_v) contributes to lifting and rotating the ladder, given by: \( F_v = F \cdot \sin(40^\circ) \)
The vertical component is critical as it creates torque about the pivot point, essential for understanding the ladder's lifting requirements.
Center of Gravity
The center of gravity is the point where the total weight of an object is thought to be concentrated. Understanding the center of gravity is necessary for analyzing how forces affect an object's balance and stability.

In the ladder scenario, the center of gravity is at its midpoint due to its uniform weight distribution, 10 meters from the pivot point. This simplifies calculations, as the gravitational force acting at this point contributes directly to the torque calculation without needing more complex integration or calculation techniques.

Knowing the center of gravity allows you to determine precisely where the gravitational force acts, thus making torque calculations and understanding object stability straightforward. It is essential for setting up correct equations for equilibrium.
Free-Body Diagram
Free-body diagrams are graphical representations used to visualize the forces acting on an object. They play a crucial role in solving static equilibrium problems by providing clarity on how forces interact.

For the ladder problem, a free-body diagram would show:
  • The gravitational force acting downward at the ladder's center of gravity.
  • The pivot point at A, where the torque is calculated.
  • The hydraulic piston at point C, with its unique angled force creating distinct horizontal and vertical components.
By using a free-body diagram, you can organize and assess the balance of forces and torques involved. This step simplifies complex physics problems and ensures that all elements are considered when solving for unknown forces like those exerted by the piston.

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Most popular questions from this chapter

The bulk modulus for bone is \(15 \mathrm{GPa}\). (a) If a diver-in-training is put into a pressurized suit, by how much would the pressure have to be raised (in atmospheres) above atmospheric pressure to compress her bones by \(0.10 \%\) of their original volume? (b) Given that the pressure in the ocean increases by \(1.0 \times 10^{4} \mathrm{~Pa}\) for every meter of depth below the surface, how deep would this diver have to go for her bones to compress by \(0.10 \%\) ? Does it seem that bone compression is a problem she needs to be concerned with when diving?

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A lead sphere has volume \(6.0 \mathrm{~cm}^{3}\) when it is resting on a lab table, where the pressure applied to the sphere is atmospheric pressure. The sphere is then placed in the fluid of a hydraulic press. What increase in the pressure above atmospheric pressure produces a \(0.50 \%\) decrease in the volume of the sphere?

A solid gold bar is pulled up from the hold of the sunken RMS Titanic. (a) What happens to its volume as it goes from the pressure at the ship to the lower pressure at the ocean's surface? (b) The pressure difference is proportional to the depth. How many times greater would the volume change have been had the ship been twice as deep? (c) The bulk modulus of lead is one- fourth that of gold. Find the ratio of the volume change of a solid lead bar to that of a gold bar of equal volume for the same pressure change.

The left-hand end of a slender uniform rod of mass \(m\) is placed against a vertical wall. The rod is held in a horizontal position by friction at the wall and by a light wire that runs from the right-hand end of the rod to a point on the wall above the rod. The wire makes an angle \(\theta\) with the rod. (a) What must the magnitude of the friction force be in order for the rod to remain at rest? (b) If the coefficient of static friction between the rod and the wall is \(\mu_{\mathrm{s}},\) what is the maximum angle between the wire and the rod at which the rod doesn't slip at the wall?

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