A ball of mass (0.5 kg) is suspended by two strings as shown in the figure at right. What is the weight of the ball? Draw a free-body diagram (FBD) for all the forces acting on the ball. (Not components.) Determine the magnitudes of the forces of tension (T1) and (T2), in the strings
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- A ball of mass (0.5 kg) is suspended by two strings as shown in the figure at right.
- What is the weight of the ball?
- Draw a free-body diagram (FBD) for all the forces acting on the ball. (Not components.)
- Determine the magnitudes of the forces of tension (T1) and (T2), in the strings.
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- NEED HELP PLEASEIn , the man has a mass of 65 kg and the crate has a mass of 110 kg. The coefficient of static friction between his shoes and the ground is μs = 0.4 and between the crate and the ground is μc = 0.3. A) Determine if the man is able to move the crate using the rope-and-pulley system shown. B) Prove your answer to part A by calculating the static frictional force F�between the man's shoes and the ground required to move the crate and the maximum static frictional force Fmax which can be developed. Express your answers in newtons to three significant figures separated by a com_ma.Block 1 (9 kg) is located on the surface of a table. A hand pushes horizontally to the right on block 1 with a normal force of 108 N. The coefficient of kinetic friction between the block and the surface equals 0.8.On a sheet of paper, draw the free body diagram for block 1 using the two-subscript notation from class. After completing the free body diagram, enter below each force and its x & y-components. Remember that the x-component is the "i" component and the y-component is the "j" component.FORCES on BLOCK 1Weight force on block 1 by Earth W1E = i + j N Normal force on block 1 by Surface N1S = i + j N Normal force on block 1 by Hand N1H = i + j N Frictional force on block 1 by Surface f1S = i + j N What is the acceleration a of block 1?a = i + j m/s2
- Consider the system in the picture below: a cart of mass M with a static friction coefficient u is connected through a massless string to a hanging mass m. M is a capital letter, m is lower case. Write them as such, or vour equations will be confusing, M We want to find the maximum value of the hanging mass m such that the system is in equilibrium. 1. Free body diagram (FBD): Draw a FBD for each: the Cart and the hanging mass. Clearly show all the forces. 2. Clearly write the equilibrium equations for the cart in the horizontal and vertical direction. 3. Clearly write the equilibrium equation for the hanging mass. 4. Solve the system of the three equations above for the hanging mass m. Show your calculation to get credit. 5. What would happen if mass m exceeds this value? Explain.e1 A double incline is setup with two ramps as shown. The left block has a mass m1, and the right block has a mass m2. The left ramp is rough with kinetic friction coefficient uy and an angle 01. The ramp on the right is frictionless with an angle 02. The pulley is massless and frictionless. Assume the system starts accelerating to the right on initial release. a) In the Space below draw a set of free body diagrams (or a single one if taking that method which is fine) to fully describe all forces in this problem. b) Find an equation for the acceleration of the system.Problem 1: A block with a mass of m rests on a frictionless surface and is subject to two forces acting on it. The first force is directed in the negative x-direction. The second acts on the body at an angle 0 measured from horizontal, as shown. If necessary, use Fs and Fk for the forces of static and kinetic friction. L. F Please use the interactive area below to draw the Free Body Diagram for this block, assuming it is in static equilibrium. FBD : Fn, F1, Fg, F2, Fs, Fk, a, v 45, 0, 90, 135, 180, 225, 270, 315, 0 Force Labels: Angle Labels: m.
- The previous step focused on friction for horizontal surfaces. As we discussed above, angled surfaces like the wedge below have different considerations. FN = mg cos(0) F₁ = mg F₁ = ? F₁ = ? Fy=mg cos(0) Specifically, we have the following considerations. 1. The normal force is not equal to mg. 2. There is a force acting along the angle of the wedge due to gravity. What is the coefficient of static friction ? Fx = mg sin(0) Let's consider the implications of this type of problem with the following examples. For all of the examples, the box has a mass of 25 kg. Imagine the box is on a surface that allows you to adjust the incline angle, 0. You find the maximum angle where the box remains stationary. That is, at any higher angle, and the box begins to slide. The static friction force at this maximum angle is 150 N. What is the angle 0 (in degrees)? Now you raise the incline to an angle of 60° greater than the angle of maximum static friction found above. At this angle, the box slides…Find I and J values and please answer all parts of this question 3. Block 1 (10 kg) is setting on the floor of an elevator. The elevator is moving upward with an increasing speed and the magnitude of the acceleration is 2 m/s2. On a sheet of paper, draw the free body diagram for block 1 using the two-subscript notation from class. After completing the free body diagram, enter below each force and its x & y-components. (use g = 10 m/2) Remember that the x-component is the "i" component and the y-component is the "j" component. -NET force on Block 1 (round to nearest integer) Fnet1 = ——- i + ——- j N The DIRECTION of the net force on block 1 is in the same direction as its acceleration. If the elevator is moving upward with an increasing speed, then the direction of the block's acceleration is up (positive y-value). The MAGNITUDE of the net force on block 1 equals its mass times acceleration, which are both given in the problem. -FORCES on BLOCK 1 Weight force on block 1…