A 3400 kg elevator is supported by a steel cable. If the tension in the cable is

Question

A 3400 kg elevator is supported by a steel cable. If the tension in the cable is 22000 N, then find the acceleration of the elevator. Use "up" as the positive direction. Two forces are applied to an object as shown. Using F_1 = 96_N, phi_1 = 68 degree, F_2 = 190_N, and phi_2 = 22 degree, find the net force on the object. magnitude direction A(n) 6.0 kg mass rests on a frictionless table. A string from this mass goes across a pulley and is attached to a(n) 10 kg mass hanging over the side of the table. Find the tension in the string and the acceleration of each mass. (Answer with the magnitude of the acceleration for either mass.) tension acceleration A 25 kg object is sliding across a surface. If its initial speed is 8.0 m/s and if the coefficient of kinetic friction between the object and the surface is 0.080, then how far will the object move before coming to rest?

Explanation / Answer

here,

mass of the elevator , m = 3400 kg

tension , t = 22000 N

let the accelration be a

t - m*g = m*a

22000 - 3400*9.8 = 3400*a

a = - 3.329 m/s^2

the accelration of the elevator is b.- 3.329 m/s^2

----------------------------

F1 = 96 * ( - cos(68) i + sin(68)) j

F1 = - 35.96 N i + 89 N j

F2 = 190 * ( cos(22) i - sin(22) j)

F2 = 176.16 N i - 71.18 N j

Fnet = F1 = F2

Fnet = 140.2 N i + 17.82 N j

magnitude , F = sqrt( 140.2^2 + 17.82^2)

F = 141.33 N

theta = arctan(17.82/140.2)

theta = 7.249 degree

the magnitude of force is c. 141.3 N

the direction is d.7.249 degree

----------------------------

m1 = 6 kg

m2 = 10 kg

the accelration , a = ( m2*g)/( m1+ m2)

a = 10 * 9.8 /( 10 + 6)

a = 6.125 m/s^2

tension in the string be t

for mass m1

t = m1*a

t = 36.75 N

the tension is b.36.75 N

the accelration is f. 6.125 m/s^2

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