[Total: 1: This free-body diagram shows three forces that act on a stone hanging at rest from two strings. tension in string 1 1.00 N tension in string 2 60° 0.58 N 30° horizontal weight of stone Figure 4.31 Calculate the horizontal component of the tension in each string. State why these two components are equal in magnitude? Calculate the vertical component of the tension in each string. a b Use your answer to part b to calculate the weight of the stone

[Total: 1: This free-body diagram shows three forces that act on a stone hanging at rest from two strings. tension in string 1 1.00 N tension in string 2 60° 0.58 N 30° horizontal weight of stone Figure 4.31 Calculate the horizontal component of the tension in each string. State why these two components are equal in magnitude? Calculate the vertical component of the tension in each string. a b Use your answer to part b to calculate the weight of the stone

Physics for Scientists and Engineers

10th Edition

ISBN:9781337553278

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter5: The Laws Of Motion

Section: Chapter Questions

Problem 18P: A bag of cement whose weight is Fg hangs in equilibrium from three wires as shown in Figure P5.18....

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3. Consider the three forces acting on the box. F1= 20.0N at an angle θ1= 45° measured from the horizontal F2= 5.00 N at an angle θ2= 60° and the weight W= 15.0 N, pulling directly down in the y- direction. What is the net force acting on a box?
1. Given that sinθ = 2/3 and cosθ = √5/3, find the cotθ, and tanθ. 2. To move a large crate across a rough floor, you push on it with a force F at an angle of 21° below the horizontal, as shown in Figure 6–18. Find the force necessary to start the crate moving, given that the mass of the crate is 32 kg and the coefficient of static friction between the crate and the floor is 0.57. Find the acceleration of the crate if the applied force is 330 N and the coefficient of kinetic friction is 0.45.
10 ) What is the resultant of two forces, F1= 100N; 60˚ and F2 = 200 N; 30˚, acting on a body? a) 200 N, 50˚ b) 290 N, 40˚ c) 300 N, 30˚ d) 305 N, 25˚
40. Analyze the system of forces on the 1.0-kg mass below and determine the acceleration. A. 2.0 m/s , 150° counterclockwise from +x-axisB. 4.5 m/s , 30° clockwise from –x-axisC. 8.0 m/s , 161° counterclockwise from +x-axisD. 16 m/s , 19° clockwise from –x-axis
1. How do you find the tension? Explain pls :)
One 3.6-kgkg paint bucket is hanging by a massless cord from another 3.6-kgkg paint bucket, also hanging by a massless cord. a) If the two buckets are pulled upward with an acceleration of 1.25 m/s^2 by the upper cord, calculate the tension in the top cord. b) If the two buckets are pulled upward with an acceleration of 1.25 m/s^2 by the upper cord, calculate the tension in the bottom cord.
** I am confused on this being below the horizontal. Is that making Fy a negative? If so, would that then affect the calculation for normal force? A box (M = 4 kg) is resting on a frictionless horizontal surface. Then there is a 25 Newton force acting onit at an angle 35 degrees below the horizontal.1. Find the horizontal component of the force, Fx - I have Fcostheta so 25cos35 = 20.479 N2. Find the vertical component of the force, Fy - I don't know if this should be 25sin35 or 25sin(-35) 3. Find the normal force acting on the box - Fn = mg - Fy (but I do not understand if this is going to be a positive or negative Fy, obviously negative would make this mg + Fy)4. Find the acceleration of the box5. Find the distance it will travel in 3 seconds.
6. A 3160 kg elevator is supported by a steel cable. If the tension in the cable is 58600_N, then find the acceleration of the elevator. Use "up" as the positive direction.
QUESTION 1 A magnet of mass 5.01 kg is suspended from the ceiling by a cord as shown. A large magnet is somewhere off to the right, pulling on the small hanging magnet with a constant force of F = 187.4 N. At what angle theta ?with respect to the vertical does the magnet hang? A 54.0 degrees B 34.4 degrees C 75.3 degrees D 14.7 degrees
(II) Two snowcats in Antarctica are towing a housing unit north, as shown in Fig. 4–50. The sum of the forces F→A and F→B exerted on the unit by the horizontal cables is north, parallel to the line L, and FA = 4500 N. Determine FB and the magnitude of F→A+F→B.
A persons head is bent over a book. The head weighs P=37.2 N and is supported by the muscle force Fm exerted by the neck extensor muscles and by the contact force Fc excerted at the atlantooccipital joint. Given that the magnitude of Fm is 60.0 N and is directed 35.0 degree below the horizontal. Find the direction of the contact force Fc. If the direction is below the horizontal, enter a negative value.
Consider the 52.0-kg mountain climber in Figure 5.22. (a) Find the tension in the rope and the force that the mountain climber must exert with her feet on the vertical rock face to remain stationary. Assume that the force is exerted parallel to her legs. Also, assume negligible force exerted by her arms. (b) What is the minimum coefficient of friction between her shoes and the cliff? Figure 5.22 Part of the climber's weight is supported by her rope and part by friction between her feet and the rock face.