2.0 kgpush45°Problem 4: (a) The 2 kg wood box in the Figure above on the left slides up a vertical wood wall whileyou push on it at a 45° angle. The coefficient of kinetic friction for wood on wood is Hk = 0.2. Whatmagnitude force Fpush should you apply to cause the box to slide up at a constant speed? Recall thatconstant speed means a = 0 m/sec?. Answer: 34.6 N.(b) Let's re-do part (a), but with the box now sliding down the vertical wood wall while you push on itat a 45° angle. Hk is still Hk = 0.2. What magnitude force Fpush should you apply to cause the box to slidedown at a constant speed? Recall that constant speed means a = 0 m/sec?. Answer: 23.1 N.Problem 5: The Figure shows two blocks connected by a cord (of negligible mass) that passes over africtionless pulley (also of negligible mass). The arrangement is known as Atwood's machine. One blockhas mass mi = 1.3 kg; the other has mass m2 = 2.8 kg. What are: (a) The magnitude of the blocks'acceleration. Answer: 3.6 m/sec?. (b) The tension in the cord? Answer: 17.4 N.Problem 6: A box with mass m = 10 kg moves on a ramp that is inclined at an angle of 0 = 55° abovethe horizontal. The coefficient of kinetic friction between the box and the ramp surface is 4 = 0.3.(a) Calculate the magnitude of the acceleration of the box if you push it down the ramp with a constantforce F = 120 N that is parallel to the ramp surface, as shown in the left in the figure below.Answer: 18.3 m/sec?.(b) Calculate the magnitude of the acceleration of the box if you push it up the ramp with a constant forceF = 120 N that is parallel to the ramp surface, as shown in the right in the figure below.Answer: 2.3 m/sec?.FmmF

Since you have asked multiple questions in a single request, we would be answering only the first…

2.0 kg push 45° Problem 4: (a) The 2 kg wood box in the Figure above on the left slides up a vertical wood wall while you push on it at a 45° angle. The coefficient of kinetic friction for wood on wood is = 0.2. What magnitude force Fpush should you apply to cause the box to slide up at a constant speed? Recall that constant speed means a = 0 m/sec2. Answer: 34.6 N.

2.0 kg push 45° Problem 4: (a) The 2 kg wood box in the Figure above on the left slides up a vertical wood wall while you push on it at a 45° angle. The coefficient of kinetic friction for wood on wood is = 0.2. What magnitude force Fpush should you apply to cause the box to slide up at a constant speed? Recall that constant speed means a = 0 m/sec2. Answer: 34.6 N.

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

See similar textbooks

Similar questions

DD=25
A uniform ladder 10m long and weighing 350N rests against a smooth wall at an angle of 30° with the wall. A 700N man stand 6m up from the bottom of the ladder. Find the horizontal force necessary at the base to keep the ladder from slipping. Draw a FBD
A rope is hung over a stationary drum with weights on each end (100N; 400N). The rope is just on the point of slipping. What is the coefficient of friction, and wrapped number. And also, if we take the same rope and making 14 turns around a stationary horizontal drum to support a heavy weight. If the coefficient of friction is Almost the same, what weight can be supported by exerting a 50-lb force at the other end of the rope?
Please, the time is short, and they respond to me as soon as possible. Please, quickly.
You need to move a cubic crate across a rough floor. You pull the crate horizontally using a rope with force F, as shown in Figure 1. The weight of the crate is 700 lb:, its density is 180 lbm/fts, and the coefficient of static friction between the crate and the floor is 0.57. Write an expression for the maximum static friction force fmax (in lb:) as a function of the mass m of the crate. Use Excel to plot fmax(m). Calculate the average pressure of the crate on the floor, in psi. Calculate the minimum force F necessary for the crate to start moving, in lb;. F Figure 1. Moving a crate on a slope.
I need to draw a free body diagram and the solution m2=88kg
suppose, the last three digit of my id :085
You pull on a crate using a rope as image, except the rope is at an angle of 20.0 ∘ above the horizontal. The weight of the crate is 325 N, and the coefficient of kinetic friction between the crate and the floor is 0.230. What must be the tension in the rope to make the crate move at a constant velocity? What is the normal force that the floor exerts on the crate?
A carnival ride has people stand inside a vertical cylinder with their backs to the wall. The cylinder starts spinning and the riders find that they are “stuck” to the wall and don’t slide down, even if the floor is removed. The ride has a radius of r. The person has a mass of m and is moving with a constant speed of v. The coefficient of static friction between the person and the wall is μs , and kinetic friction μk. The person is only touching the wall, not touching the floor. What is the magnitude of the acceleration of the person? In what direction does it point? The speed is constant. Why is the acceleration not zero? Briefly explain. No equations!
Please answer and show how you got it You are on an elevator fixed to a wall that can move up and down freely while you pull with a force of 150 N on a massless rope (the orange line in the diagram below) with length L. This rope is connected to one end of a beam with the same length, L, and mass m=20 kg. The left end of the beam is in contact with the same wall the elevator is on and the coefficient of static friction =0.2 between the beam and the wall. See diagram below for a representation of the situation and note that free fall acceleration points directly downward 0 L Part A) Label all the forces acting on the beam and discuss the direction of any torques exerted on the beam about the contact point between the beam and the wall Part B) What is the maximum angle the beam can subtend with respect to the horizontal _m such that it will just begin to slip? Potentially useful trig identities are sin(2x)=2sin(x)cos, sin(180-x)=sin(x), sin(90+x)=cos(x)
I neee typing clear urjent i will give 5 upvotes
You construct a version of the cart and bucket, but with a slope whose angle can be adjusted. You use a cart of mass 167 kg and a bucket of mass 57.0 kg. The cable has negligible mass, and there is no friction. What must be the angle of the slope so that the cart moves downhill at a constant speed and the bucket moves upward at the same constant speed? With this choice of angle, what will be the tension in the cable?