In the figure, a nonuniform bar is suspended at rest in a horizontal position by two massless cords as shown in the figure here. One cord makes the angle e = 20.5° with the vertical; the other makes the angle p = 69.5° with the vertical. If the length L of the bar is 5.1 m, compute the distance x from the left end of the bar to its center of mass. -1 com X = i
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Rotational Equilibrium And Rotational Dynamics
In physics, the state of balance between the forces and the dynamics of motion is called the equilibrium state. The balance between various forces acting on a system in a rotational motion is called rotational equilibrium or rotational dynamics.
Equilibrium of Forces
The tension created on one body during push or pull is known as force.
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- The larger ball in the picture has a mass m1 = 30 kg while the smaller ball has a mass m2 = 10 kg. (a) If the larger ball is placed 1 m away from the fulcrum, how far away from the fulcrum must the smaller ball be placed in order for the seesaw to be in equilibrium? Here the fulcrum is in the middle of the seesaw. (b) Now the fulcrum is located one fourth of the way from the left end of the plank and the balls are placed at either end. If the larger ball has the same mass as in part (a), what must the mass of the smaller ball be in order for the seesaw to be in equilibrium? Ignore the mass of the plank.A uniform ladder stands on a rough floor and rests against a frictionless wall as shown in the figure. Since the floor is rough, it exerts both a normal force N1 and a frictional force f1 on the ladder. However, since the wall is frictionless, it exerts only a normal force N2 on the ladder. The ladder has a length of L = 4.1 m, a weight of WL = 61.0 N, and rests against the wall a distance d = 3.75 m above the floor. If a person with a mass of m = 90 kg is standing on the ladder, determine the following. (a) the forces exerted on the ladder when the person is halfway up the ladder (Enter the magnitude only.) N1 = N N2 = N f1 = N (b) the forces exerted on the ladder when the person is three-fourths of the way up the ladder (Enter the magnitude only.) N1 = N N2 = N f1 = NA uniform ladder stands on a rough floor and rests against a frictionless wall as shown in the figure. Since the floor is rough, it exerts both a normal force N1 and a frictional force f1 on the ladder. However, since the wall is frictionless, it exerts only a normal force N2 on the ladder. The ladder has a length of L = 4.4 m, a weight of WL = 53.5 N, and rests against the wall a distance d = 3.75 m above the floor. If a person with a mass of m = 90kg is standing on the ladder, determine the following. (a) the forces exerted on the ladder when the person is halfway up the ladder (Enter the magnitude only.) N1 = ? N N2 = ? N f1 = ? N (b) the forces exerted on the ladder when the person is three-fourths of the way up the ladder (Enter the magnitude only.) N1 = ? N N2 = ? N f1 = ? N
- A beam resting on two pivots has a length of L = 6.00 m and mass M = 91.0 kg.The pivot under the left end exerts a normal force n1 on the beam, and the second pivot placed a distance ℓ = 4.00 m from the left end exerts a normal force n2. A woman of mass m = 62.0 kg steps onto the left end of the beam and begins walking to the right as in the figure below. The goal is to find the woman's position when the beam begins to tip. (b) Where is the woman when the normal force n1 is the greatest?x = m(c) What is n1 when the beam is about to tip? N(d) Use the force equation of equilibrium to find the value of n2 when the beam is about to tip. N(e) Using the result of part (c) and the torque equilibrium equation, with torques computed around the second pivot point, find the woman's position when the beam is about to tip.x = m(f) Check the answer to part (e) by computing torques around the first pivot point.x = mWhy is the following situation impossible? A uniform beam of mass mb = 3.00 kg and length ℓ = 1.00 m supports blocks with masses m1 = 5.00 kg and m2 = 15.0 kg at two positions as shown. The beam rests on two triangular blocks, with point P a distance d = 0.300 m to the right of the center of gravity of the beam. The position of the object of mass m2 is adjusted along the length of the beam until the normal force on the beam at O is zero.A beam resting on two pivots has a length of L = 6.00 m and mass M = 88.0 kg. The pivot under the left end exerts a normal force n1 on the beam, and the second pivot placed a distance ℓ = 4.00 m from the left end exerts a normal force n2. A woman of mass m = 61.0 kg steps onto the left end of the beam and begins walking to the right as in the figure below. The goal is to find the woman's position when the beam begins to tip. Where is the woman when the normal force n1 is the greatest? X=0 c. (c) What is n1 when the beam is about to tip? 0 (d) Use the force equation of equilibrium to find the value of n2 when the beam is about to tip. (e) Using the result of part (c) and the torque equilibrium equation, with torques computed around the second pivot point, find the woman's position when the beam is about to tip (f) Check the answer to part (e) by computing torques around the first pivot point x =
- the figure shows a horizontal human arm lifting a dumbbell. the forearm is in equilibrium under the action of the weight w of the dumbbell, the tension t in the tendon connected to the biceps muscle, and the force e exerted on the forearm by the upper arm at the elbow joints. we neglect the weight of the forearm itself. (for clarity, the point a where the tendon is attached is drawn farther from the elbow than its actual position.) given the weight w and the angle between the tension force and the horizontal, find t and the two components of ē (three unknown scalar quantities in all)A beam resting on two pivots has a length of L = 6.00 m and mass M = 92.0 kg. The pivot under the left end exerts a normal force n1 on the beam, and the second pivot placed a distance ℓ = 4.00 m from the left end exerts a normal force n2. A woman of mass m = 62.5 kg steps onto the left end of the beam and begins walking to the right as in the figure below. The goal is to find the woman's position when the beam begins to tip. (c) Using the (force equation of equilibrium to find the value of n2 when the beam is about to tip). and the torque equilibrium equation, with torques computed around the second pivot point, find the woman's position when the beam is about to tip.x = z____m(d) Check the answer to part (c) by computing torques around the first pivot point.x = ____m(e) Except for possible slight differences due to rounding, is the answer the same? Yes or NoA beam resting on two pivots has a length of L = 6.00 m and mass M = 94.0 kg. The pivot under the left end exerts a normal force n1 on the beam, and the second pivot placed a distance ℓ = 4.00 m from the left end exerts a normal force n2. A woman of mass m = 51.5 kg steps onto the left end of the beam and begins walking to the right as in the figure below. The goal is to find the woman's position when the beam begins to tip. (e) Using the result of part c and the torque equilibrium equation, with torques computed around the second pivot point, find the woman's position when the beam is about to tip. (part c answer: n1= 0 when the beam is about to tip_x = _____m(f) Check the answer to part (e) by computing torques around the first pivot point.x = ____mExcept for possible slight differences due to rounding, is the answer the same? Yes or No
- A beam resting on two pivots has a length of L = 6.00 m and mass M = 94.0 kg. The pivot under the left end exerts a normal force n1 on the beam, and the second pivot placed a distance ℓ = 4.00 m from the left end exerts a normal force n2. A woman of mass m = 51.5 kg steps onto the left end of the beam and begins walking to the right as in the figure below. The goal is to find the woman's position when the beam begins to tip. (a)Where is the woman when the normal force n1 is the greatest? x = _____m(b) What is n1 when the beam is about to tip?____N(c) Use the force equation of equilibrium to find the value of n2 when the beam is about to tip.____NA 10.0-kg monkey climbs a uniform ladder with weight w = 1.09 ✕ 102 N and length L = 3.40 m as shown in the figure below. The ladder rests against the wall at an angle of ? = 60°. The upper and lower ends of the ladder rest on frictionless surfaces, with the lower end fastened to the wall by a horizontal rope that is frayed and that, can support a maximum tension of only 80.0 N.a nonuniform bar is suspended at rest in a horizontal position by two massless cords. One cord makes the angle u=36.9° with the vertical; the other makes the angle f =53.1° with the vertical. If the length L of the bar is 6.10 m, compute the distance x from the left end of the bar to its center of mass.