Understanding the Force on a Car on a Curve

The question describes how the force needed to prevent a car from skidding on a curve is related to several factors: the car's weight, its speed, and the radius of the curve. This relationship is given by a variation equation.

We are told that the force (\(F\)) varies directly with the weight (\(W\)) and the square of the speed (\(v^2\)), and inversely with the radius (\(r\)). This can be written as:

\[ F \propto \frac{W v^2}{r} \]

To turn this proportionality into an equation, we introduce a constant of proportionality, let's call it \(k\):

\[ F = k \frac{W v^2}{r} \]

Our goal is to first find the value of this constant \(k\) using the information from the first scenario, and then use \(k\) to calculate the force needed for the second scenario.

Calculating the Proportionality Constant

From the first scenario, we are given:

• Force (\(F_1\)) = 6125 pound-force
• Weight (\(W_1\)) = 2750 pounds
• Speed (\(v_1\)) = 35 mph
• Radius (\(r_1\)) = 550 feet

We can plug these values into our equation to solve for \(k\):

\[ 6125 = k \frac{2750 \times 35^2}{550} \]

First, calculate \(35^2\):

\[ 35^2 = 1225 \]

Now substitute this back into the equation:

\[ 6125 = k \frac{2750 \times 1225}{550} \]

Let's simplify the term \(\frac{2750}{550}\). \(2750 \div 550 = 5\).

\[ 6125 = k \times 5 \times 1225 \]

\[ 6125 = k \times (5 \times 1225) \]

\[ 6125 = k \times 6125 \]

Now, solve for \(k\):

\[ k = \frac{6125}{6125} \]

\[ k = 1 \]

The proportionality constant \(k\) is 1.

Finding the Required Force for the Second Scenario

Now we know the relationship is simply \(F = \frac{W v^2}{r}\). For the second scenario, we are given:

• Weight (\(W_2\)) = 3600 pounds
• Speed (\(v_2\)) = 50 mph
• Radius (\(r_2\)) = 750 feet

We need to find the force (\(F_2\)) required to keep this car from skidding. Using the equation with \(k=1\):

\[ F_2 = 1 \times \frac{W_2 v_2^2}{r_2} \]

\[ F_2 = \frac{3600 \times 50^2}{750} \]

First, calculate \(50^2\):

\[ 50^2 = 2500 \]

Substitute this back into the equation:

\[ F_2 = \frac{3600 \times 2500}{750} \]

Now, perform the multiplication in the numerator:

\[ 3600 \times 2500 = 9,000,000 \]

So, the equation becomes:

\[ F_2 = \frac{9,000,000}{750} \]

Finally, divide to find \(F_2\):

\[ F_2 = 12000 \]

The required force is 12000 pound-force.

Summary of Calculations

The force required to keep the 3600 pound car, travelling at 50 mph, from skidding on a curve of radius 750 feet is 12000 pound-force.

Revision Table: Understanding Variation Problems

Additional Information: Physics of Skidding on a Curve

The force discussed in this problem is related to the centripetal force required to make the car turn. When a car turns on a flat road, the centripetal force is primarily provided by the static friction force between the tires and the road. The formula \(F = m \frac{v^2}{r}\) is the standard physics formula for centripetal force, where \(m\) is mass and \(v\) is velocity (speed), and \(r\) is the radius. Since weight (\(W\)) is mass (\(m\)) times acceleration due to gravity (\(g\)), \(W = mg\), we can write \(m = W/g\). Substituting this into the centripetal force formula gives \(F = \frac{W}{g} \frac{v^2}{r}\). Comparing this to our variation equation \(F = k \frac{W v^2}{r}\), we see that the proportionality constant \(k\) is related to \(1/g\), scaled by units. Our calculation found \(k=1\), which suggests that the units used in the problem (pound-force, pounds, mph, feet) must incorporate the gravitational constant in such a way that \(k\) becomes 1. In practice, the maximum static friction force depends on the coefficient of static friction and the normal force (which is equal to the weight on a flat road). The car will skid if the required centripetal force exceeds the maximum possible static friction force.