Concept explainers
Consider the three connected objects shown in Figure P5.88. Assume first that the inclined plane is friction-less and that the system is in equilibrium. In terms of m, g, and θ, find (a) the mass M and (b) the tensions T, and T2. Now assume that the value of Af is double the value found in part (a). Find (c) the acceleration of each object and (d) the tensions T1 and T2. Next, assume that the coefficient of static friction between m and 2m and the inclined plane is m, and that the system is in equilibrium. Find (e) the maximum value of M and (0 the minimum value of M. (g) Compare the values of T2 when M has its minimum and maximum values.
(a)
The expression for the mass of
Answer to Problem 5.88AP
The expression for the mass
Explanation of Solution
The free body diagram of the three connected objects is shown in Figure below,
Figure (1)
From Figure (1), the equilibrium forces acts on the object of mass
Here,
From the Newton’s second law of motion, the net force on the object of mass
Here,
Substitute
From Figure (1), the equilibrium forces act on the object of mass
Here,
From the Newton’s second law of motion, the net force on the object of mass
Substitute
Add the equation (1) with equation (2).
From Figure (1), the equilibrium forces act on the object of mass
Here,
Substitute
From the Newton’s second law of motion, the net force on the object of mass
Substitute
The system is in equilibrium so value of the acceleration is zero.
Substitute
Conclusion:
Therefore, the expression for the mass
(b)
The expressions for tensions
Answer to Problem 5.88AP
The expression for the tension
Explanation of Solution
From part (a), the expression for the mass of
From part (a), the equilibrium forces act on the object is,
The system is in equilibrium, the value of acceleration is zero so the net force acts on the system is also zero.
Substitute
Substitute
Thus, the expression for the tension
From part (a), the equation (2) is,
Substitute
Substitute
Thus, the expression for the tension
Conclusion:
Therefore, the expression for the tension
(c)
The acceleration of each object.
Answer to Problem 5.88AP
The acceleration of each object is
Explanation of Solution
Given info: The value of mass
From part (a), the expression for the mass when it is double represents as,
From part (a), the equation (1) is,
Rearrange the above equation.
From part (a), the equation (2) is,
Rearrange the above equation.
Substitute
From Figure (1), the equilibrium forces act on the object of mass
Substitute
Subtract the equation (3) from equation (4).
Conclusion:
Therefore, the acceleration of each object is
(d)
The expressions for tensions
Answer to Problem 5.88AP
The expression for the tension
Explanation of Solution
From part (c), the expression for the acceleration is,
From part (c), the equation for tension
Substitute
Thus, the expression for tension
From part (c), the equation (3) is,
Substitute
Thus, the expression for tension
Conclusion:
Therefore, the expression for the tension
(e)
The maximum value of
Answer to Problem 5.88AP
The maximum value of
Explanation of Solution
Given info: The coefficient of static friction between mass
The static friction forces on the masses
Figure (2)
From the Figure (2), the normal force on the object of mass
The expression for the static friction force on the object of mass
Here,
Substitute
From the Figure (1), the equilibrium forces acts on the object of mass
Substitute
Rearrange the above equation for
Substitute
From the Figure (2), the normal force on the object of mass
Substitute
From the Figure (1), the equilibrium forces acts on the object of mass
Substitute
Add equation (4) with the above equation.
Substitute
The equilibrium forces on the block of mass
Substitute
Conclusion:
Therefore, the maximum value of
(f)
The minimum value of
Answer to Problem 5.88AP
The minimum value of
Explanation of Solution
From part (e), the expression of tension
The equilibrium forces acts on block for minimum mass of
Substitute
Conclusion:
Therefore, the minimum value of
(g)
The difference between the tension
Answer to Problem 5.88AP
The difference between the tension for maximum and minimum mass is
Explanation of Solution
From part (e), the expression for maximum mass is,
From part (f), the expression for the minimum mass is,
From part (e), the expression for the tension for maximum mass is,
From part (f), the expression for the tension for maximum mass is,
Compare both the above equation.
Substitute
Conclusion:
Therefore, the difference between the tension for maximum and minimum mass is
Want to see more full solutions like this?
Chapter 5 Solutions
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
- A heavy chandelier with mass 125 kg is hung by chains in equilibrium from the ceiling of a concert hall as shown in Figure P5.77, with 1 = 37.0 and 2 = 64.0. Assuming the chains are massless, what are the tensions FT1, FT2, and FT3 in the three chains? FIGURE P5.77arrow_forwardTwo objects, m1 = 3.00 kg and m2 = 8.50 kg, are attached by a massless cord passing over a frictionless pulley as shown in Figure P5.51. Assume the horizontal surface is frictionless. a. Draw a free-body diagram for each of the two objects. b. What is the tension in the cord? c. What is the magnitude of the acceleration of the two objects? FIGURE P5.51 Problems 51 and 65.arrow_forward(a) What is the minimum force of friction required to hold the system of Figure P4.74 in equilibrium? (b) What coefficient of static friction between the 100.-N block and the table ensures equilibrium? (c) If the coefficient of kinetic friction between the 100.-N block and the table is 0.250, what hanging weight should replace the 50.0-N weight to allow the system to move at a constant speed once it is set in motion? Figure P4.74arrow_forward
- Two blocks connected by a rope of negligible mass are being dragged by a horizontal force (Fig. P5.13). Suppose F = 68.0 N, m1 = 12.0 kg, m2 = 18.0 kg, and the coefficient of kinetic friction between each block and the surface is 0.100. (a) Draw a free-body diagram for each block. Determine (b) the acceleration of the system and (c) the tension T in the rope. Figure P5.13arrow_forwardInitially, the system of objects shown in Figure P5.49 is held motionless. The pulley and all surfaces and wheels are frictionless. Let the force F be zero and assume that m1 can move only vertically. At the instant after the system of objects is released, Find (a) the tension T in the string, (b) the acceleration of m2, (c) the acceleration of M, and (d) the acceleration of m1. (Note: The pulley accelerates along with the cart.) Figure P5.49 Problems 49 and 53arrow_forwardAn object of mass m1 = 5.00 kg placed on a frictionless, horizontal table is connected to a string that passes over a pulley and then is fastened to a hanging object of mass m2 = 9.00 kg as shown in Figure P5.22. (a) Draw free-body diagrams of both objects. Find (b) the magnitude of the acceleration of the objects and (c) the tension in the string. Figure P5.22 Problems 22 and 29.arrow_forward
- Three objects are connected on a table as shown in Figure P5.14. The coefficient of kinetic friction between the block of mass m2 and the table is 0.350. The objects have masses of m1 = 4.00 kg, m2 = 1.00 kg, and m3 = 2.00 kg, and the pulleys are frictionless. (a) Draw a free-body diagram of each object. (b) Determine the acceleration of each object, including its direction. (c) Determine the tensions in the two cords. What If? (d) If the tabletop were smooth, would the tensions increase, decrease, or remain the same? Explain. Figure P5.14arrow_forwardTwo blocks are connected by a rope that passes over a massless and frictionless pulley as shown in Figure P5.41. Given that m0 = 15.93 kg and m2 = 10.45 kg, determine the magnitudes of the tension in the rope and the blocks acceleration. FIGURE P5.41arrow_forwardAn object of mass m1 = 5.00 kg placed on a frictionless, horizontal table is connected to a string that passes over a pulley and then is fastened to a hanging object of mass m2 = 9.00 kg as shown in Figure P4.28. (a) Draw free-body diagrams of both objects. Find (b) the magnitude of the acceleration of the objects and (c) the tension in the string. Figure P4.28arrow_forward
- Two blocks, each of mass m, are hung from the ceiling of an elevator as in Figure P4.33. The elevator has an upward acceleration a. The strings have negligible mass. (a) Find the tensions T1 and T2 in the upper and lower strings in terms of m, a, and g. (b) Compare the two tensions and determine which string would break first if a is made sufficiently large. (c) What are the tensions if the cable supporting the elevator breaks? Figure P4.33 Problems 33 and 34.arrow_forwardAll object of mass m = 500 kg is suspended from the ceiling of an accelerating truck as shown in Figure P6.21. Taking a = 3.00 m/s2, find (a) the angle 0 that the string makes with the vertical and (b) the tension T in the string.arrow_forwardA 9.00-kg hanging object is connected by a light, inextensible cord over a light, frictionless pulley to a 5.00-kg block that is sliding on a flat table (Fig. P5.7). Taking the coefficient of kinetic friction as 0.200, find the tension in the string. Figure P5.7arrow_forward
- Physics for Scientists and Engineers, Technology ...PhysicsISBN:9781305116399Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningPrinciples of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningPhysics for Scientists and EngineersPhysicsISBN:9781337553278Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning
- Physics for Scientists and Engineers with Modern ...PhysicsISBN:9781337553292Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningCollege PhysicsPhysicsISBN:9781305952300Author:Raymond A. Serway, Chris VuillePublisher:Cengage LearningCollege PhysicsPhysicsISBN:9781285737027Author:Raymond A. Serway, Chris VuillePublisher:Cengage Learning