What is the angle B if the acceleration is g/2? Answer 26.6 degrees SOLUTIONSET UP Our free-body diagram is shown in Figure 5.7b. The forcesacting on the key are its weight w = mg and the string tension T. Wedirect the x axis to the right (in the direction of the acceleration) and they 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, butthe car, string, and key are all accelerating in the +x direction. Thus,there must be a horizontal component of force acting on the key, andso we use Newton's second law in component form: Fx = max andΣF, = may.We find the components of the string tension, as shown inFigure 5.7b. The sum of the horizontal components of force isΣF = T sinß,and the sum of the vertical components isΣΕ, = T cosB + (-mg).The x component of the acceleration of the car, string, and key isax = a, and the y component of acceleration is zero, soEXAMPLEEF=Tsinß= max,EF, = T cosß+ (-mg) = may = 0.5.2 Applications of Newton's Second LawRearranging terms, we obtainT sinB = max129T cos ß = mg.When we divide the first equation by the second, we getarom indurax = 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 iszero; when ß = 45°, ax = g.REFLECT We note that ß can never be 90° because that would requirean infinite acceleration. We note also that the relationship between axand 3 doesn't depend on the mass of the key, but it does depend on theacceleration due to gravity.Practice Problem: What is the angle ß if the acceleration is g/2?Answer: 26.6°. you pushA. The normal force from the floor is less than the UB. 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'tmove vertically. Therefore, the vertical component of the accelerationof the box must be zero. This means that the upward-pointing normalEXAMPLE 5.5 A simple accelerometerIn 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 theother end of the string (Figure 5.7a). A protractor taped to the light allows you to measure the angle thestring makes with the vertical. Your friend drives the car while you make measurements. When the car has aconstant acceleration with magnitude a toward the right, the string hangs at rest (relative to the car), makingan angle B with the vertical. (a) Derive an expression for the acceleration a in terms of the mass m and themeasured angle B. (b) In particular, what is a when ß = 45°? When ß = 0?à(a) Low-tech accelerometerA FIGURE 5.7A FIGURE 5.6àBIMass mT cos BB05290T sin BW = mg(b) Free-body diagramfor the keyVideo Tutor SolusCONTIN

Question

What is the angle B if the acceleration is g/2?  Answer 26.6 degrees SOLUTIONSET UP Our free-body diagram is shown in Figure 5.7b. The forcesacting on the key are its weight w = mg and the string tension T. Wedirect the x axis to the right (in the direction of the acceleration) and they 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, butthe car, string, and key are all accelerating in the +x direction. Thus,there must be a horizontal component of force acting on the key, andso we use Newton's second law in component form: Fx = max andΣF, = may.We find the components of the string tension, as shown inFigure 5.7b. The sum of the horizontal components of force isΣF = T sinß,and the sum of the vertical components isΣΕ, = T cosB + (-mg).The x component of the acceleration of the car, string, and key isax = a, and the y component of acceleration is zero, soEXAMPLEEF=Tsinß= max,EF, = T cosß+ (-mg) = may = 0.5.2 Applications of Newton's Second LawRearranging terms, we obtainT sinB = max129T cos ß = mg.When we divide the first equation by the second, we getarom indurax = 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 iszero; when ß = 45°, ax = g.REFLECT We note that ß can never be 90° because that would requirean infinite acceleration. We note also that the relationship between axand 3 doesn't depend on the mass of the key, but it does depend on theacceleration due to gravity.Practice Problem: What is the angle ß if the acceleration is g/2?Answer: 26.6°. you pushA. The normal force from the floor is less than the UB. 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'tmove vertically. Therefore, the vertical component of the accelerationof the box must be zero. This means that the upward-pointing normalEXAMPLE 5.5 A simple accelerometerIn 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 theother end of the string (Figure 5.7a). A protractor taped to the light allows you to measure the angle thestring makes with the vertical. Your friend drives the car while you make measurements. When the car has aconstant acceleration with magnitude a toward the right, the string hangs at rest (relative to the car), makingan angle B with the vertical. (a) Derive an expression for the acceleration a in terms of the mass m and themeasured angle B. (b) In particular, what is a when ß = 45°? When ß = 0?à(a) Low-tech accelerometerA FIGURE 5.7A FIGURE 5.6àBIMass mT cos BB05290T sin BW = mg(b) Free-body diagramfor the keyVideo Tutor SolusCONTIN

Accepted Answer

What is the angle B if the acceleration is g/2?

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