Concept explainers
(a)
The acceleration of each of the blocks.
(a)
Explanation of Solution
Introduction:
The free-body diagram for both the block is shown in Figure 1
Figure 1
In the above diagram is shown that the force on the block
Conclusion:
Therefore, the free body diagram for both blocks is shown in Figure 1
(b)
The expression for the acceleration of the block and the tension in the string.
(b)
Answer to Problem 82P
The expression for the acceleration is
Explanation of Solution
Calculation:
The equation of motion for the block
The equation of motion for the block
The expression for the acceleration of the block is evaluated as,
The expression for the tension in the string is evaluated a,
Conclusion:
Therefore, the expression for the acceleration is
(c)
Whether the expressions for the tension and the acceleration gives the plausible results if
(c)
Answer to Problem 82P
The value of tension for the condition
Explanation of Solution
Calculation:
For the value of
For the value of
For the value of
For the value of
For the value of
For the value of
Conclusion:
Therefore, the value of tension for the condition
Want to see more full solutions like this?
Chapter 4 Solutions
PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS
- Consider a setup with a horizontal table, 2 masses, a string, and a massless frictionless pulley. Assume on the table m1= 1.8kg and hanging m2= 1.2kg. Also assume the coefficent of friction, uk = 0.20 on the table. a) For each mass, identify the forces, draw a free body diagram, and set up equations for Newton's second law. You can also write an equation of the system. b) Find the accleration after release.arrow_forwardThe force F is applied to a 60-1b box, always directed down an angle of 0 - 30° from the horizontal as shown. The magnitude of the force is increased until the boxbegins slide. Use the static coefficient of friction us = 0.6 and the kinetic coefficient offriction uk = 0.3.Sketch a free body diagram and the kinetic diagram of the boxFind the required F that can cause motion of the boxDetermine the box's initial accelerationarrow_forwardAn Atwood's machine is a simply two masses connected by a string that runs over a pulley, so that there is a mass on either side of the pulley. If one of the masses is 12.5 kg and the other mass is 9.75 kg, then determine the acceleration of the masses and the tension in the string. For simplicity you can make the usual assumptions about the string and the pulley. Include a force (or free-body) diagram of the situation.arrow_forward
- In the system shown in the figure, mA = 2 kg and mB = 5 kg.the mass is connected to each other by rope. A 12 N Fforce is applied to body A at an angle of 10 °.There is only friction between body B and the inclined planeand the kinetic coefficient of friction is 0.3.a) Draw the free body diagram of each object.b) Find the acceleration of the system.arrow_forwardDraw a diagram of the situation and label the givens in the problem: The cart's mass and the hanger's mass. Also label the direction of the cart and hanger's acceleration. Draw two force diagrams. One for the cart and one for the hanger. You can treat the track as frictionless. Remember that the force in the string will be the same throughout the entire string. Then define an x-y coordinate system and find the x and y components of the forces in your diagrams.arrow_forwardTwo boxes with masses m1= -60.0 kg and m2= -30.0 kg are connected by a light string that passes over a light, frictionless pulley One box rests on a frictionless ramp that rises at 40.0° above the horizontal, and the system is released form rest. (a) Make a force (or free-body) diagram of each of the boxes and include the coordinate system used for each box. (b) Write down Newton's 2ad law for each box in each relevant direction. (c) What is the magnitude of the tension in the string, and the magnitude of the acceleration of the boxes?arrow_forward
- The diagram below shows an object of mass m= 20kg being acted upon by a force F=50N at an angle of θ=30o. The coefficient of kinetic friction between the object and the surface is 0.100. Draw a Free Body diagram for the object. Be sure to show all forces as well as the reference x,y axis and by Using Newton’s 2nd law, write the force equations for both the x and the y directions. Write these equations in terms of the letters m, F, θ, FN, g, f (little f stands for the friction force). X-direction: _______________________________________________________________ Y-direction:________________________________________________________________ b) Solve for the Normal Force c)Use your answer from part C to find the frictional force. Please answer all three partsarrow_forwardA) The students draw the following Free Body Diagram for the situation. Is thế Freë Body Diagram correct? Check the directions of all of the forces and make sure that the students didn't miss any forces. Fix any mistakes that you find. B) The students next wrote the following statement for Newton's 2nd Law for the system: Fn 2Mg T EF system = (2M + M)a =F +FN- 2Mg - T +T – Mg . Is this correct? Fi: mistakes that the students made and explain why they were mistakes. C) A second student noticed that this equation can't be solved for Tensi Instead, they wrote the following equation for Block 2M: EF M Block 2M = 0 = luded thatT must be 40N Is this correct? If not, explain why not. Mgarrow_forwardPlease solve it properly and step by step process with illustrationsarrow_forward
- Draw "free body digarams" with clearly labeled vectors for: (a) The glider only (b) The hanging mass only (c) The two object system, glider plus hanging mass and clearly label each diagram. For (c), draw clearly labeled vectors only for forces that are external to the two-object system.arrow_forwardConsider the 65.0 kg Ice skater being pushed by two others shown In the figure. Frot F2 Free-body diagram F, F, F, (a) Find the direction (in degrees) and magnitude (in N) of F,et, the total force exerted on her by the others, glven that the magnitudes F, and F, are 25.2 N and 15.4 N, respectively. direction o (counterclockwise from the direction of F, is positive) magnitude (b) What is her initial acceleration (in m/s?) if she is initlally stationary and wearing steel-bladed skates that point in the direction of F? (c) What is her acceleration (in m/s2) assuming she is already moving in the direction of Fot? Remember that friction is always in the opposite direction of motion or attempted motion between surfaces in contact. m/s? (in the direction of Frot)arrow_forwardShown is a free-body diagram. For the problem, (a) redraw the free-body diagram and (b) write a short description of a real object for which this is the correct free-body diagram.arrow_forward
- University Physics Volume 1PhysicsISBN:9781938168277Author:William Moebs, Samuel J. Ling, Jeff SannyPublisher:OpenStax - Rice University