Concept explainers
(a)
The magnitude of contact force between the two masses.
(a)

Answer to Problem 79P
The magnitude of contact force between the two masses is
Explanation of Solution
Write the expression for the Newton’s second law of motion.
Here,
If the plane is frictionless, then the only force on the two masses are the normal forces and the weight with the frictional forces equalling zero. Since there is no force directed up the incline, the normal force between the two blocks must be zero.
Conclusion:
Therefore, the magnitude of contact force between the two masses is
(b)
The magnitude of contact force between two masses if the coefficient of kinetic friction between the masses and the plane are equal.
(b)

Answer to Problem 79P
The magnitude of contact force between two masses if the coefficient of kinetic friction between the masses and the plane are equal is
Explanation of Solution
Write the expression for the Newton’s law for the mass
Here,
Write the expression for the Newton’s law for the mass
Here,
Use equation (III) to write the expression for
Here,
Use equation (III) in (II) to solve for
Write the expression for the Newton’s law for the mass
Here,
Write the expression for the Newton’s law for the mass
Here,
Use equation (VI) to write the expression for
Here,
Use equation (VII) in (V) to solve for
Write the expression for the Newton’s third law of motion.
If the normal force is non-zero, then the acceleration must be equal.
Use equation (VIII) and (IV) in (X) and it becomes,
The coefficient of kinetic friction between the masses and the plane are equal.
Use equation (XII) in (XI) to solve for the contact forces between the two masses.
The only way this can be true with
Conclusion:
Therefore, the magnitude of contact force between two masses if the coefficient of kinetic friction between the masses and the plane are equal is
(c)
If
(c)

Answer to Problem 79P
The magnitude of contact force between the two blocks is
Explanation of Solution
Use equation (IX) and (XI) to solve for
Conclusion:
Substitute
Therefore, the magnitude of contact force between the two blocks is
Want to see more full solutions like this?
Chapter 4 Solutions
Bundle: College Physics: Reasoning And Relationships, 2nd + Webassign Printed Access Card For Giordano's College Physics, Volume 1, 2nd Edition, Multi-term
- dry 5. (a) When rebuilding her car's engine, a physics major must exert 300 N of force to insert a c piston into a steel cylinder. What is the normal force between the piston and cyli=030 What force would she have to exert if the steel parts were oiled? k F = 306N 2 =0.03 (arrow_forwardInclude free body diagramarrow_forwardInclude free body diagramarrow_forward
- Test 2 МК 02 5. (a) When rebuilding her car's engine, a physics major must exert 300 N of force to insert a dry = 0.03 (15 pts) piston into a steel cylinder. What is the normal force between the piston and cylinder? What force would she have to exert if the steel parts were oiled? Mk Giren F = 306N MK-0.3 UK = 0.03 NF = ?arrow_forward2. A powerful motorcycle can produce an acceleration of 3.50 m/s² while traveling at 90.0 km/h. At that speed the forces resisting motion, including friction and air resistance, total 400 N. (Air resistance is analogous to air friction. It always opposes the motion of an object.) What force does the motorcycle exert backward on the ground to produce its acceleration if the mass of the motorcycle with rider is 245 ke? a = 350 m/s 2arrow_forward2. A powerful motorcycle can produce an acceleration of 3.50 m/s² while traveling at 90.0 km/h. At that speed the forces resisting motion, including friction and air resistance, total 400 N. (Air resistance is analogous to air friction. It always opposes the motion of an object.) What force does the motorcycle exert backward on the ground to produce its acceleration if the mass of the motorcycle with rider is 245 kg? (10 pts) a = 3.50 m/s 2 distance 90 km/h = 3.50m/62 M = 245garrow_forward
- Using Table 17-4, determine the approximate temperature of metal that has formed a dark blue color.arrow_forwardA positively charged disk has a uniform charge per unit area σ. dq R P x The total electric field at P is given by the following. Ek [2 - x (R² + x2) 1/2 Sketch the electric field lines in a plane perpendicular to the plane of the disk passing through its center.arrow_forwardConsider a closed triangular box resting within a horizontal electric field of magnitude E = 8.02 104 N/C as shown in the figure below. A closed right triangular box with its vertical side on the left and downward slope on the right rests within a horizontal electric field vector E that points from left to right. The box has a height of 10.0 cm and a depth of 30.0 cm. The downward slope of the box makes an angle of 60 degrees with the vertical. (a) Calculate the electric flux through the vertical rectangular surface of the box. kN · m2/C(b) Calculate the electric flux through the slanted surface of the box. kN · m2/C(c) Calculate the electric flux through the entire surface of the box. kN · m2/Carrow_forward
- The figure below shows, at left, a solid disk of radius R = 0.600 m and mass 75.0 kg. Tu Mounted directly to it and coaxial with it is a pulley with a much smaller mass and a radius of r = 0.230 m. The disk and pulley assembly are on a frictionless axle. A belt is wrapped around the pulley and connected to an electric motor as shown on the right. The turning motor gives the disk and pulley a clockwise angular acceleration of 1.67 rad/s². The tension T in the upper (taut) segment of the belt is 145 N. (a) What is the tension (in N) in the lower (slack) segment of the belt? N (b) What If? You replace the belt with a different one (one slightly longer and looser, but still tight enough that it does not sag). You again turn on the motor so that the disk accelerates clockwise. The upper segment of the belt once again has a tension of 145 N, but now the tension in the lower belt is exactly zero. What is the magnitude of the angular acceleration (in rad/s²)? rad/s²arrow_forwardA bridge truss extends x = 217 m across a river (shown in the figure below) where 0 = 40°. The structure is free to slide horizontally to permit thermal expansion. The structural components are connected by pin joints, and the masses of the bars are small compared with the mass of a 1300 kg car at the center. Calculate the force of tension or compression in each structural component (in N). B D T T T T T 22820 AB AC BC ||| || || || BD N ---Select--- N ---Select--- N ---Select--- N ---Select--- DE N ---Select--- T DC= N ---Select--- TEC N ---Select--- с ✓ Earrow_forwardno ai pleasearrow_forward
- Physics for Scientists and Engineers: Foundations...PhysicsISBN:9781133939146Author:Katz, Debora M.Publisher:Cengage LearningPrinciples of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningClassical Dynamics of Particles and SystemsPhysicsISBN:9780534408961Author:Stephen T. Thornton, Jerry B. MarionPublisher:Cengage Learning
- College PhysicsPhysicsISBN:9781305952300Author:Raymond A. Serway, Chris VuillePublisher:Cengage LearningCollege PhysicsPhysicsISBN:9781285737027Author:Raymond A. Serway, Chris VuillePublisher:Cengage LearningPhysics for Scientists and Engineers with Modern ...PhysicsISBN:9781337553292Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning





