A vaulter holds a 25.2-N pole in equilibrium by exerting an upward force with he
ID: 2242059 • Letter: A
Question
A vaulter holds a 25.2-N pole in equilibrium by exerting an upward force with her leading hand and a downward force with her trailing hand as shown in the figure below. Point C is the center of gravity of the pole. (Let L1 = 2.25 m, L2 = 1.55 m, and L3 = 0.700 m.) What is the magnitude of ? What is the magnitude of ? A wheel starts from rest and rotates with constant angular acceleration to reach an angular speed of 12.8 rad/s in 2.98 s. Find the magnitude of the angular acceleration of the wheel. Find the angle in radians through which it rotates in this time interval. Find the net torque on the wheel in the figure below about the axle through O, taking a = 9.00 cm and b = 23.0 cm. (Assume that the positive direction is counterclockwise.) A 1 510-kg automobile has a wheel base (the distance between the axles) of 3.30 m. The automobile's center of mass is on the centerline at a point 1.05 m behind the front axle. Find the force exerted by the ground on each wheel.Explanation / Answer
1)
2) a) alfa = W2-W1/t = 12.8-0/2.98 = 4.295 s
b) theta = 0.5*alfa*t^2 = 0.5*4.295*2.98^2 = 19.071 radians
3)
4) The torque from the Front wheels must balance the torque of the weight of the vehicle.
The weight of the vehicle is given as 1.05 m from the front wheels, that is 3.3 - 1.05 = 2.25 m from the rear wheels.
Since ? = FL, we get
Ffront*3.3 = 1530*9.8*2.25
Ffront = 10223 N
For the rear wheels, we can use the sum of the forces. The total weight of the car must be supported by the two front and two rear wheels.
Frear + Ffront = (1530*9.8)
Frear + 10223 = (1530*9.8)
Frear = 4771N
For the front, the two wheels support 10223 N which means each wheel supports 5111.5 N. = 5.1115 kN
For the rear, the two wheels support 4771 N which means each wheel supports 2386 N = 2.386 kN
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