Concept explainers
Compare the net force on a heavy trunk when it’s (a) at rest on the floor; (b) being slid across the floor at constant speed; (c) being pulled upward in an elevator whose cable tension equals the combined weight of the elevator and trunk; and (d) sliding down a frictionless ramp.
(a)
Answer to Problem 1FTD
Explanation of Solution
According to Newton’s first law of motion a body will remain in rest until it is acted upon by external unbalanced force. Therefore the net force acting on a body at rest is zero.
The forces acting on a body are weight of the body which is down ward and normal reaction which is upward. For a body at rest on the floor therefore the normal reaction will be equal to that of weight of the body. Thus net force will be zero.
Conclusion:
Thus, the net force on a heavy trunk when it is at rest on the floor is zero.
(b)
Answer to Problem 1FTD
Explanation of Solution
According to Newton’s first law of motion a body will remain in uniform motion until it is acted upon by external unbalanced force. Therefore the net force acting on a body in uniform motion is zero.
The forces acting on the trunk are the weight of the trunk which is acting downward and normal reaction which is acting upward, driving force along the direction of motion, air resistance opposite to the direction of force and friction opposite to the direction of motion. The normal reaction balances weight of the trunk and driving force balances the sum of air resistance and friction. Thus net force acting on a constant speed car is zero.
Conclusion:
Thus, the net force on a heavy trunk when it is being slid across the floor at constant speed is zero.
(c)
Answer to Problem 1FTD
Explanation of Solution
According to Newton’s first law of motion a body will remain in uniform motion until it is acted upon by external unbalanced force.
The forces acting on the elevator and trunk is the weight of the body acting downward and tension of the cable acting upward. Since both are same net force acting on the system is zero. Thus total acceleration is zero. Therefore net force on elevator and trunk is zero.
Conclusion:
Thus, the net force on a heavy trunk when it is being pulled upward in an elevator whose cable tension equals the combined weight of the elevator and trunk is zero
(d)
Answer to Problem 1FTD
Explanation of Solution
According to Newton’s first law of motion a body will move with acceleration only if there is any net force acting on it.
The trunk sliding down is similar to free fall. The only force is the weight of the body which is downward. Therefore net force on the trunk is downward.
Conclusion:
Thus, the net force on a heavy trunk when it is sliding down a frictionless ramp is downward.
Want to see more full solutions like this?
Chapter 5 Solutions
Essential University Physics: Volume 1 (3rd Edition)
Additional Science Textbook Solutions
Microbiology with Diseases by Body System (5th Edition)
Anatomy & Physiology (6th Edition)
Campbell Biology (11th Edition)
Introductory Chemistry (6th Edition)
Human Anatomy & Physiology (2nd Edition)
Cosmic Perspective Fundamentals
- Please solve and answer this problem correctly please. Thank you!!arrow_forwarda) Use the node-voltage method to find v1, v2, and v3 in the circuit in Fig. P4.14. b) How much power does the 40 V voltage source deliver to the circuit? Figure P4.14 302 202 w w + + + 40 V V1 80 Ω 02 ΣΑΩ 28 A V3 + w w 102 202arrow_forwardPlease solve and answer this problem correctly please. Thank you!!arrow_forward
- You're on an interplanetary mission, in an orbit around the Sun. Suppose you make a maneuver that brings your perihelion in closer to the Sun but leaves your aphelion unchanged. Then you must have Question 2 options: sped up at perihelion sped up at aphelion slowed down at perihelion slowed down at aphelionarrow_forwardThe force of the quadriceps (Fq) and force of the patellar tendon (Fp) is identical (i.e., 1000 N each). In the figure below angle in blue is Θ and the in green is half Θ (i.e., Θ/2). A) Calculate the patellar reaction force (i.e., R resultant vector is the sum of the horizontal component of the quadriceps and patellar tendon force) at the following joint angles: you need to provide a diagram showing the vector and its components for each part. a1) Θ = 160 degrees, a2) Θ = 90 degrees. NOTE: USE ONLY TRIGNOMETRIC FUNCTIONS (SIN/TAN/COS, NO LAW OF COSINES, NO COMPLICATED ALGEBRAIC EQUATIONS OR ANYTHING ELSE, ETC. Question A has 2 parts!arrow_forwardThe force of the quadriceps (Fq) and force of the patellar tendon (Fp) is identical (i.e., 1000 N each). In the figure below angle in blue is Θ and the in green is half Θ (i.e., Θ/2). A) Calculate the patellar reaction force (i.e., R resultant vector is the sum of the horizontal component of the quadriceps and patellar tendon force) at the following joint angles: you need to provide a diagram showing the vector and its components for each part. a1) Θ = 160 degrees, a2) Θ = 90 degrees. NOTE: USE DO NOT USE LAW OF COSINES, NO COMPLICATED ALGEBRAIC EQUATIONS OR ANYTHING ELSE, ETC. Question A has 2 parts!arrow_forward
- No chatgpt pls will upvotearrow_forwardThe force of the quadriceps (Fq) and force of the patellar tendon (Fp) is identical (i.e., 1000 N each). In the figure below angle in blue is Θ and the in green is half Θ (i.e., Θ/2). A) Calculate the patellar reaction force (i.e., R resultant vector is the sum of the horizontal component of the quadriceps and patellar tendon force) at the following joint angles: you need to provide a diagram showing the vector and its components for each part. a1) Θ = 160 degrees, a2) Θ = 90 degrees. NOTE: USE ONLY TRIGNOMETRIC FUNCTIONS (SIN/TAN/COS, NO LAW OF COSINES, NO COMPLICATED ALGEBRAIC EQUATIONS OR ANYTHING ELSE, ETC. Question A has 2 parts!arrow_forwardNo chatgpt pls will upvotearrow_forward
- No chatgpt pls will upvotearrow_forwardSolve and answer the question correctly please. Thank you!!arrow_forward་ The position of a particle is described by r = (300e 0.5t) mm and 0 = (0.3t²) rad, where t is in seconds. Part A Determine the magnitude of the particle's velocity at the instant t = 1.5 s. Express your answer to three significant figures and include the appropriate units. v = Value Submit Request Answer Part B ? Units Determine the magnitude of the particle's acceleration at the instant t = 1.5 s. Express your answer to three significant figures and include the appropriate units. a = Value A ? Unitsarrow_forward
- College PhysicsPhysicsISBN:9781305952300Author:Raymond A. Serway, Chris VuillePublisher:Cengage LearningCollege PhysicsPhysicsISBN:9781285737027Author:Raymond A. Serway, Chris VuillePublisher:Cengage LearningPrinciples of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning
- Physics for Scientists and Engineers: Foundations...PhysicsISBN:9781133939146Author:Katz, Debora M.Publisher:Cengage LearningGlencoe Physics: Principles and Problems, Student...PhysicsISBN:9780078807213Author:Paul W. ZitzewitzPublisher:Glencoe/McGraw-HillUniversity Physics Volume 1PhysicsISBN:9781938168277Author:William Moebs, Samuel J. Ling, Jeff SannyPublisher:OpenStax - Rice University