A. An object moves in a circle with a diameter of 8 meters at a frequency of 8 r

Question

A. An object moves in a circle with a diameter of 8 meters at a frequency of 8 rev/s. What is the period, tangential speed, and centripetal acceleration?

Period =  seconds

Tangential Speed = m/s

Centripetal Acceleration = m/s2

B. A 11 kg crate is moved from the ground up a loading platform 6.9 m long and 2.4 m high. Assume the coefficient of kinetic friction is 0.2. What is the actual mechanical advantage and incline plane efficiency?

Actual MA =

Incline Efficiency =  %

C. A 1,100 kg load is hung from a 6 meter beam protruding from a wall. If the beam has a cross sectional area of 23 cm2 and the shear modulus is 23,700 MPa. What are the shearing stress and the downward deflection of the beam?

Shearing Stress = N/m2

Downward Deflection =  meters

Explanation / Answer

Q1.

angular frequency=8 revolutions/seconds=8*2*pi radian/seconds

then frequency=angular frequency/(2*pi)=8 Hz

time period=1/frequency=0.125 seconds

tangential speed=angular frequency*radius=8*2*pi*(8/2)=201.06 m/s

centripetal acceleration=tangential speed^2/radius=201.06^2/(8/2)=1.0106*10^4 m/s^2

Q2.

length is 6.9 m and height is 2.4 m

angle with horizontal=theta=arcsin(2.4/6.9)=20.354 degrees

normal force=m*g*cos(theta)=11*9.8*cos(20.354)=101.07 N

this is the force exerted by the incline on the object.

force required to move the block up is=component of weight of the block along the incline + friction force

=m*g*sin(theta)+friction coefficient*normal force

=11*9.8*sin(20.354)+0.2*101.07=57.709 N

hence actual mechanical advantage=Fout/Fin=force exerted by the incline/force exerted by you=101.07/57.709=1.7514

actual work done=potential energy at height of 2.4 m=11*9.8*2.4=258.72 J

work done in moving 6.9 m=57.709*6.9=398.19 J

hence efficiency=258.72/398.19=64.974%

Q3.force on the beam=weight of the load=1100*9.8=10780 N

hence stress=force/area=10780/(23*10^(-4))=4.687*10^6 N/m^2

as shear modulous=force*length/(area*deflection)

==>deflection=force*length/(area*shear modulous)=10780*6/(23*10^(-4)*23700*10^6)=1.1866 mm

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