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Practice Problem: What is the angle ß if the acceleration is g/2? Answer: 26.6°.

Practice Problem: What is the angle ß if the acceleration is g/2? Answer: 26.6°.

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What is the angle B if the acceleration is g/2? Answer 26.6 degrees
SOLUTION SET UP Our free-body diagram is shown in Figure 5.7b. The forces acting on the key are its weight w = mg and the string tension T. We direct the x axis to the right (in the direction of the acceleration) and the y axis vertically upward. SOLVE Part (a): This problem may look like an equilibrium problem, but it isn't. The string and the key are at rest with respect to the car, but the car, string, and key are all accelerating in the +x direction. Thus, there must be a horizontal component of force acting on the key, and so we use Newton's second law in component form: Fx = max and ΣF, = may. We find the components of the string tension, as shown in Figure 5.7b. The sum of the horizontal components of force is ΣF = T sinß, and the sum of the vertical components is ΣΕ, = T cosB + (-mg). The x component of the acceleration of the car, string, and key is ax = a, and the y component of acceleration is zero, so EXAMPLE EF=Tsinß= max, EF, = T cosß+ (-mg) = may = 0. 5.2 Applications of Newton's Second Law Rearranging terms, we obtain T sinB = max 129 T cos ß = mg. When we divide the first equation by the second, we get arom indur ax = g tan B. The acceleration ax is thus proportional to the tangent of the angle B. Part (b): When ß = 0, the key hangs vertically and the acceleration is zero; when ß = 45°, ax = g. REFLECT We note that ß can never be 90° because that would require an infinite acceleration. We note also that the relationship between ax and 3 doesn't depend on the mass of the key, but it does depend on the acceleration due to gravity. Practice Problem: What is the angle ß if the acceleration is g/2? Answer: 26.6°.
Transcribed Image Text:SOLUTION SET UP Our free-body diagram is shown in Figure 5.7b. The forces acting on the key are its weight w = mg and the string tension T. We direct the x axis to the right (in the direction of the acceleration) and the y axis vertically upward. SOLVE Part (a): This problem may look like an equilibrium problem, but it isn't. The string and the key are at rest with respect to the car, but the car, string, and key are all accelerating in the +x direction. Thus, there must be a horizontal component of force acting on the key, and so we use Newton's second law in component form: Fx = max and ΣF, = may. We find the components of the string tension, as shown in Figure 5.7b. The sum of the horizontal components of force is ΣF = T sinß, and the sum of the vertical components is ΣΕ, = T cosB + (-mg). The x component of the acceleration of the car, string, and key is ax = a, and the y component of acceleration is zero, so EXAMPLE EF=Tsinß= max, EF, = T cosß+ (-mg) = may = 0. 5.2 Applications of Newton's Second Law Rearranging terms, we obtain T sinB = max 129 T cos ß = mg. When we divide the first equation by the second, we get arom indur ax = g tan B. The acceleration ax is thus proportional to the tangent of the angle B. Part (b): When ß = 0, the key hangs vertically and the acceleration is zero; when ß = 45°, ax = g. REFLECT We note that ß can never be 90° because that would require an infinite acceleration. We note also that the relationship between ax and 3 doesn't depend on the mass of the key, but it does depend on the acceleration due to gravity. Practice Problem: What is the angle ß if the acceleration is g/2? Answer: 26.6°.
you push A. The normal force from the floor is less than the U B. The normal force from the floor is equal to the box's weight. C. The normal force from the floor is greater than the box's weight. SOLUTION As the box slides horizontally across the floor, it doesn't move vertically. Therefore, the vertical component of the acceleration of the box must be zero. This means that the upward-pointing normal EXAMPLE 5.5 A simple accelerometer In this example we will use Newton's second law to calculate the acceleration of a key hanging in a car. You tape one end of a piece of string to the ceiling light of your car and hang a key with mass m on the other end of the string (Figure 5.7a). A protractor taped to the light allows you to measure the angle the string makes with the vertical. Your friend drives the car while you make measurements. When the car has a constant acceleration with magnitude a toward the right, the string hangs at rest (relative to the car), making an angle B with the vertical. (a) Derive an expression for the acceleration a in terms of the mass m and the measured angle B. (b) In particular, what is a when ß = 45°? When ß = 0? à (a) Low-tech accelerometer A FIGURE 5.7 A FIGURE 5.6 à BI Mass m T cos BB 05290 T sin B W = mg (b) Free-body diagram for the key Video Tutor Solus CONTIN
Transcribed Image Text:you push A. The normal force from the floor is less than the U B. The normal force from the floor is equal to the box's weight. C. The normal force from the floor is greater than the box's weight. SOLUTION As the box slides horizontally across the floor, it doesn't move vertically. Therefore, the vertical component of the acceleration of the box must be zero. This means that the upward-pointing normal EXAMPLE 5.5 A simple accelerometer In this example we will use Newton's second law to calculate the acceleration of a key hanging in a car. You tape one end of a piece of string to the ceiling light of your car and hang a key with mass m on the other end of the string (Figure 5.7a). A protractor taped to the light allows you to measure the angle the string makes with the vertical. Your friend drives the car while you make measurements. When the car has a constant acceleration with magnitude a toward the right, the string hangs at rest (relative to the car), making an angle B with the vertical. (a) Derive an expression for the acceleration a in terms of the mass m and the measured angle B. (b) In particular, what is a when ß = 45°? When ß = 0? à (a) Low-tech accelerometer A FIGURE 5.7 A FIGURE 5.6 à BI Mass m T cos BB 05290 T sin B W = mg (b) Free-body diagram for the key Video Tutor Solus CONTIN
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