A spring and block are in the arrangement of the figure. When the block is pulled out to x = +4.0 cm, we must apply a force of magnitude 370N to hold it there. We pull the block to x = 11.0 cm and then release it. How much work does the spring do on the block when the block moves from x, = +5.0 cm to (a) x = +4.0 cm, (b) x = -4.0 cm, (c) x= -5.0 cm, and (d) x = -10.0 cm? x=0 Block F-0 attached to spring ellee (a) x positive F, negative
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- An eye bolt is used to attach 3 cables to a steel plate. The tension in the three cables create F,=200 Ibf, F2=250 lbf, and F3=100 lbf with 0 = 30 degrees and p=27.1 degrees. If the eye bolt is in equilibrium, what is the x-component of the sum of other forces (reaction force) on the bolt? If you add up the three force vectors, the reaction force you are looking for will just be in the opposite direction to put the eye bolt in equilibrium. The x-direction is positive and to the right. Eye bolt rsteel plate Nut & WasherThe following diagram corresponds to questions 1 to 3. A beam rests on two sharp edges as shown on the image. The beam has a length of 7,50 m and a mass of 4,25 kg. Object 1 has a mass of 1,50 kg; object 2 has a mass of 2,35 kg. Point P is 0,50 m form the center of the beam. d 2 Pl CG 1. How much is the torque done by object 1 around point P? A. 55,2 N*m B. 7,36 N*m С. 62,5 N*m D. Object 1 doesn't exert a torque.A spring and block are in the arrangement of the figure. When the block is pulled out to x- +5.0 cm, we must apply a force of magnitude 370 N to hold it there. We pull the block tox-11.0 cm and then release it. How much work does the spring do on the block when the block moves from x₁ +5.0 cm to (a) x +3.0 cm, (b)x= -3.0 cm, (c) x- -5.0 cm, and (d) x - -9.0 cm? (a) Number i 7.2 (b) Number i (c) Number 0 (d) Number 5.76 -25.2 x=0) E-0 00000000 x positive F, negative ooo Fromos Units Units J Units J J Units J (a) (b) -Block attached to spring x negative F, positive
- Question #3: In the figure below, a 100 N weight flower pot is suspended by three ropes. The tension in rope AB is three times the tension in ropes AC and AD. Find the dimension “d” on the figure which corresponds to the z-coordinate of point B. Also find the tensions in ropes AB, AC, and AD. Please do each step as I am trying to check my answers. I got to the point of setting the ijk components equal to zero. I'm stuck at multiplying the tension in vector AB times 3.The figure shows a person wearing weight boots and doing lower leg flexion/extension exercise in a sitting position to strengthen the quadriceps muscles and a simple mechanical model of his leg. W1 is the weight of the lower leg, W0 is the weight of the boot, the magnitude of the pulling force applied to the tibia by the quadriceps muscles through the FM patellar tendon, the magnitude of the reaction force acting on the FJ tibiofemoral joint. Point O is the center of the tibiofemoral joint, point A is the point where the patellar tendon attaches to the tibia, point B is the center of gravity of the lower leg, point C is the center of gravity of the weight boot. The distances between point O and points A, B and C were measured as a=12 cm, b=24 cm and c=36 cm, respectively. The angle that the long axis of the tibia makes with the horizontal is β=39°, the angle between the line of action of the quadriceps muscle strength and the long axis of the tibia is α=16°. Points O, A, B and C lie…The figure shows a person wearing weight boots and doing lower leg flexion/extension exercise in a sitting position to strengthen the quadriceps muscles and a simple mechanical model of his leg. W1 is the weight of the lower leg, W0 is the weight of the boot, the magnitude of the pulling force applied to the tibia by the quadriceps muscles through the FM patellar tendon, the magnitude of the reaction force acting on the FJ tibiofemoral joint. Point O is the center of the tibiofemoral joint, point A is the point where the patellar tendon attaches to the tibia, point B is the center of gravity of the lower leg, point C is the center of gravity of the weight boot. The distances between point O and points A, B and C were measured as a=12 cm, b=24 cm and c=36 cm, respectively. The angle that the long axis of the tibia makes with the horizontal is β=39°, the angle between the line of action of the quadriceps muscle strength and the long axis of the tibia is α=16°. Points O, A, B and C lie…
- The figure shows a person wearing weight boots and doing lower leg flexion/extension exercise in a sitting position to strengthen the quadriceps muscles and a simple mechanical model of his leg. W1 is the weight of the lower leg, W0 is the weight of the boot, the magnitude of the pulling force applied to the tibia by the quadriceps muscles through the FM patellar tendon, the magnitude of the reaction force acting on the FJ tibiofemoral joint. Point O is the center of the tibiofemoral joint, point A is the point where the patellar tendon attaches to the tibia, point B is the center of gravity of the lower leg, point C is the center of gravity of the weight boot. The distances between point O and points A, B and C were measured as a=12 cm, b=24 cm and c=36 cm, respectively. The angle that the long axis of the tibia makes with the horizontal is β=39°, the angle between the line of action of the quadriceps muscle strength and the long axis of the tibia is α=16°. Points O, A, B and C lie…The figure shows a person wearing weight boots and doing lower leg flexion/extension exercise in a sitting position to strengthen the quadriceps muscles and a simple mechanical model of his leg. W1 is the weight of the lower leg, W0 is the weight of the boot, the magnitude of the pulling force applied to the tibia by the quadriceps muscles through the FM patellar tendon, the magnitude of the reaction force acting on the FJ tibiofemoral joint. Point O is the center of the tibiofemoral joint, point A is the point where the patellar tendon attaches to the tibia, point B is the center of gravity of the lower leg, point C is the center of gravity of the weight boot. The distances between point O and points A, B and C were measured as a=13 cm, b=27 cm and c=36 cm, respectively. The angle that the long axis of the tibia makes with the horizontal is β=34°, the angle between the line of action of the quadriceps muscle strength and the long axis of the tibia is α=18°. Points O, A, B and C lie…The figure shows a person wearing weight boots and doing lower leg flexion/extension exercise in a sitting position to strengthen the quadriceps muscles and a simple mechanical model of his leg. W1 is the weight of the lower leg, W0 is the weight of the boot, the magnitude of the pulling force applied to the tibia by the quadriceps muscles through the FM patellar tendon, the magnitude of the reaction force acting on the FJ tibiofemoral joint. Point O is the center of the tibiofemoral joint, point A is the point where the patellar tendon attaches to the tibia, point B is the center of gravity of the lower leg, point C is the center of gravity of the weight boot. The distances between point O and points A, B and C were measured as a=12 cm, b=24 cm and c=36 cm, respectively. The angle that the long axis of the tibia makes with the horizontal is β=39°, the angle between the line of action of the quadriceps muscle strength and the long axis of the tibia is α=16°. Points O, A, B and C lie…
- a) Let's start with a simplified model where both M, and M₂ are applied perpendicular to the moment arm. The moment arm lengths are d₁ = 3 cm, d₂ 15 cm, and d3 = 34 cm. M₁ M₂ da W Calculate the ratio of the mechanical advantage of the brachioradialis(M₂) to the mechanical advantage of the biceps(M₁). Ratio of Mechanical Advantage MA2 -) = 5 MA1An eye bolt is used to attach 3 cables to a steel plate. The tension in the three cables create F1=200 lbf, F2=250 lbf, and F3=100 lbf with 0 = 30 degrees and p=24.1 degrees. If the eye bolt is in equilibrium, what is the x-component of the sum of other forces on the bolt (force from the nut and plate on the bolt) ? If you add up the three force vectors, the sum other force you are looking for will just be in the opposite direction to put the eye bolt in equilibrium. The x-direction is positive to the right. For example, if you find the sum of forces 1, 2, and 3 are 100 Ibf to the right, then the other forces in the x-direction must be pointing to the left (-100 lbf) to put the eye bolt in equilibrium. Eye bolt 2steel 3 plate Nut t WonsherConsider the system shown below. Answer Questions 6-11 based on this figure. How many degrees-of-freedom does the system shown below has? The two pulleys have the identical mass, radius, and the moment of inertia. т, 1, R k т, 1, R Variables x: vertical position of the moving pulley 8: rotation of the moving pulley p: rotation of the fixed pulley m: pulley mass R: pulley radius I: mass moment of the pulley about its own center of mass. k: spring stiffness T: cable tension Assumptions 1. x = 0, 0 = 0, and o = 0 when the spring is undeformed. 2. The cables and springs have negligible mass. 3. There is no slip between the pulley and the cable. 4. Only two-dimensional planar motion is allowed.