Problem 50 - College Physics

Question

Two people carry a heavy electric motor by placing it on a light board 2.00 m long. One person lifts at one end with a force of \(400 \mathrm{N},\) and the other lifts the opposite end with a force of 600 \(\mathrm{N}\) . (a) What is the weight of the motor, and where along the board is its center of gravity located? (b) Suppose the board is not light but weighs \(200 \mathrm{N},\) with its center of gravity at its center, and the two people each exert the same forces as before. What is the weight of the motor in this case, and where is its center of gravity located?

Step-by-Step Solution

Answer

(a) 1000 N, 1.2 m; (b) 800 N, 1.25 m

Step 1: Define Variables and Understand the Problem

We have a board 2.00 m long, with one person lifting one end with a force of 400 N, and the other lifting the opposite end with a force of 600 N. We need to find the weight of the electric motor when the board is light (a) and when it weighs 200 N itself (b). We also need to find the center of gravity in both cases.

Step 2: Calculate Total Force in Case A

In case (a), since the board is light, the total force exerted by the two people (400 N and 600 N) should equal the weight of the motor.\[ W_{ ext{motor}} = F_1 + F_2 = 400 ext{ N} + 600 ext{ N} = 1000 ext{ N} \]

Step 3: Determine Location of Center of Gravity in Case A

Let the center of gravity of the motor be at a distance \( x \) from the end where 400 N is applied. The torques about the 400 N end must equal:\[ 600 ext{ N} \times 2.00 ext{ m} = W_{ ext{motor}} \times x \]Substituting the value of \( W_{ ext{motor}} \),\[ 600 imes 2 = 1000 \times x \]Solving for \( x \) gives:\[ x = \frac{1200}{1000} = 1.2 ext{ m} \]

Step 4: Calculate Added Force Components in Case B

Now the board has a weight of 200 N, so we add this to the system. Distribute the board's weight equally as its center of gravity is at its midpoint (1.00 m). Thus, the total forces become:\[ F_{ ext{total}} = 400 ext{ N} + 600 ext{ N} = 1000 ext{ N} \] but now this includes the board.

Step 5: Determine Motor's Weight in Case B

Since the board weighs 200 N, the force due to the motor is:\[ W_{ ext{motor}} = F_{ ext{total}} - W_{ ext{board}} \]\[ W_{ ext{motor}} = 1000 ext{ N} - 200 ext{ N} = 800 ext{ N} \]

Step 6: Find Center of Gravity in Case B

Using torque equilibrium around the 400 N lift point:\[ 600 imes 2.00 - 200 imes 1.00 = W_{ ext{motor}} \times x \]\[ 1200 - 200 = 800 \times x \]\[ 1000 = 800 \times x \]\[ x = \frac{1000}{800} = 1.25 ext{ m} \]

Step 7: Conclusion

The motor's weight in case (a) is 1000 N with its center of gravity at 1.2 m from the 400 N end. In case (b), the motor's weight is 800 N with its center of gravity at 1.25 m from the 400 N end.

Key Concepts

Center of Gravity

The center of gravity is a crucial concept in physics, especially when dealing with objects in equilibrium. It is the point where the total weight of an object appears to act. For simple shapes, the center of gravity often lies at the geometric center. However, for irregular shapes or complex systems like our motor on the board, it requires calculation.
In our exercise, the center of gravity of the motor determines the distribution of forces needed to keep the board in equilibrium. The location of the center of gravity changes based on whether the board is light or has its own weight, influencing how the weight of the motor is perceived and distributed.
Knowing the center of gravity is essential for solving problems in statics, as it helps predict how forces need to be applied to keep the system balanced.

Statics

Statics is the branch of mechanics concerned with objects in a state of rest or moving at constant velocity. In statics, we analyze forces without considering movement or changes in velocity. Through this exercise, we learn that an object is in equilibrium when the total sum of forces and the sum of torques are zero.
In the first scenario, the motor and board system reaches equilibrium when the lifting forces sum up to the weight of the motor, and the torques about any point balance out. This understanding allows us to accurately calculate the motor's weight and center of gravity.
By considering torques, students can see how equilibrium is not just about balancing forces but also involves balancing moments around a pivot point. Successfully applying these statics principles is key in physics problem solving, particularly in structural engineering and mechanical design.

Physics Problem Solving

Physics problem solving requires a methodical approach to dissect complex systems into more manageable parts. In this exercise, we break the problem into smaller tasks: defining variables, calculating forces, and assessing torque balance. This allows us to find the solutions step-by-step.
For students, understanding how to isolate forces and assess their impact on equilibrium is vital. It starts with identifying known values and manipulating equations to discover unknowns. This exercise demonstrates the utility of setting equations relative to a pivot, using algebra to solve for distances, and ultimately determining weights and positions.

• Identify what is known and unknown in the problem.
• Apply Newton's laws to set up equations for force and torque.
• Use algebra to solve these equations and find the solution.
• Check work for accuracy to ensure all forces and moments are balanced.

Following these steps builds problem-solving abilities applicable beyond physics.

Newton's Laws

Newton's laws lay the foundation for understanding mechanics, including torque and equilibrium problems. The most relevant law in this context is Newton's First Law, which states that an object will remain at rest or in uniform motion unless acted upon by a net external force.
In our exercise, we see Newton's laws at work in statics. The forces applied by the people lifting the board must counterbalance the weight of the motor and the board itself to maintain equilibrium. Here, Newton’s laws guide us on how forces interact to cause or prevent motion.
Applying these laws, the exercise demonstrates how the motor's weight and placement affect the balance. The sum of the forces, as well as the sum of the torques (which are the rotational equivalent of linear forces), must equal zero for the system to remain stationary. Understanding these principles helps in analyzing stability and ensuring structures serve their intended function safely.