Answered: A non-uniform rod is horizontal, and… | bartleby

Related questions

Question

A non-uniform rod is horizontal, and has one end at x = 0
and the other end at æ
20 m. In order to keep the rod
horizontal, and above the ground, vertical forces Fo = 5 N (at
x = 0) and FL = 3 N (at x = 20 m) need to be applied at the ends.
What is the weight W of the rod, and what is the x-coordinate of
the center of gravity of the rod?

Expert Solution

Trending now

This is a popular solution!

Step by step

Solved in 2 steps with 2 images

SEE SOLUTION Check out a sample Q&A here

Blurred answer

Similar questions

A man holds a 198-N ball in his hand, with the forearm horizontal (see the figure). He can support the ball in this position because of the flexor muscle force M, which is applied perpendicular to the forearm. The forearm weighs 18.5 N and has a center of gravity as indicated. Find (a) the magnitude of M and the (b) magnitude and (c) direction (as a positive angle counterclockwise from horizontal) of the force applied by the upper arm bone to the forearm at the elbow joint. Upper arm bone - Flexor muscle M Elbow joint cg 0.0510 mt T0.0890 m 0.330 m-
Ea- d ms A horizontal L=1.4 m long mb3D12kg uniform bar is hinged on the left end and pulled at the right end by a cable. The cable makes 30° angle with horizontal. A 22 kg store sign is suspended below the bar at d=0.18 m from the right end. Find the magnitude of horizontal hinge force.
6m T 10m A 8m Figure 4 A B 6. A horizontal beam AB of mass m = 20kg is supported from end B by a cable and hinged to a A vertical wall at end A. Calculate the magnitude of the tension T in the cable when Batman, who has a mass M = 80 kg, stands midway along the beam. What are the x and y components of the force at the hinge?
The center of gravity of Kelly's outstretched arm is located 21 cm from the axis of rotation in her shoulder. Her arm has a mass of 4.2 kg and is 56 cm long. Assuming the deltoid's force is horizontal, what is the magnitude of the force her deltoid muscle must exert on Kelly's upper arm so she can hold a 6.0 kg mass stationary 15 degree above the horizontal as shown? Her deltoid muscle attaches to her upper arm at a point 18 cm from her shoulder joint. 6 kg 15° 21 cm Fdeltiod
Page 4 of 4 Torque And Static Equilibrium 4. A 5-m uniform ladder of mass 15 kg is held stationary against a frictionless wall. If the ladder is in a state of impending motion when it makes an angle of 65° with respect to the floor, calculate the force exerted by the wall onto the ladder. Solution and Answer: Since the ladder is uniform, then the center of gravity of the ladder is at its geometric center. We use the first and second conditions for equilibrium in order to tackle this problem. The chosen axis of rotation is at the point where the ladder makes contact with the ground. Let Fwall and f be the forces due to the wall and due to friction, respectively. From the first conditions of equilibrium, we get: force exerted by wall weighe ermal force ction =f+(-Fwatt) = 0 B = n + (-W) = 0 The second condition for equilibrium is: Twall + (-Triction) + Tn + (-Tweight) = 0 The frictional force and the normal force are located at the pivot point. This further simplifies to: Twall +…
A ladder is leaning against a vertical wall, and both ends of the ladder are at the point of slipping. The coefficient of friction between the ladder and the horizontal surface is ?1=0.165 and the coefficient of friction between the ladder and the wall is ?2=0.153 . Determine the maximum angle α with the vertical that the ladder can make without falling on the ground.
A man holds a 195-N ball in his hand, with the forearm horizontal (see the figure). He can support the ball in this position because of the flexor muscle force M , which is applied perpendicular to the forearm. The forearm weighs 20.7 N and has a center of gravity as indicated. Find (a) the magnitude of M and the (b) magnitude and (c) direction (as a positive angle counterclockwise from horizontal) of the force applied by the upper arm bone to the forearm at the elbow joint. Upper arm bone- Flexor muscle M Elbow cg joint 0.0510 m+ 0.0890 m -0.330 m- (a) Number i Units (b) Number i Units (c) Number i Units >
H 60° Figure 6 G M 8. A uniform rod HG of length 2L m and mass m = 2 kg is hinged at end H, as shown in Figure 6 A mass Mg=100 N is hung at the other end G. A horizontal cable at the midpoint of the rod holds it at an angle of 60° to the horizontal. The cable is under a constant tension T. (a) Draw a diagram showing all the forces acting on the rod and determine the tension T in the cable. (b) Calculate both the x and y components (Fr and F₁) of the force F at the hinge H. (c) Will the net force F at H act along the rod HG? Justify your answer.
A man holds a 179-N ball in his hand, with the forearm horizontal (see the figure). He can support the ball in this position because of the flexor muscle force , which is applied perpendicular to the forearm. The forearm weighs 20.3 N and has a center of gravity as indicated. Find (a) the magnitude of M and the (b) magnitude and (c) direction (as a positive angle counterclockwise from horizontal) of the force applied by the upper arm bone to the forearm at the elbow joint. Upper arm bone Flexor muscle M Elbow joint 0.0510 mt 0.0890 m -0.330 m- (a) Number Units (b) Number Units (c) Number Units