pic 3 (20 points) (i) A flatbed truck slowly tilts its bed upward to dispose of

Question

pic 3 (20 points)

(i) A flatbed truck slowly tilts its bed upward to dispose of a 95.0kg crate. For small

angles of tilt the crate stays put, but when the tilt angle exceeds23.2o

, the crate

begins to slide. What is the coefficient of static friction between the bed of the truck

and the crate? (4 points)

(ii) Suppose the coefficients of static and kinetic friction between the crate and the truck

bed are 0.415 and 0.382, respectively. (a) Does the crate begin to slide at a tilt angle

that is greater than, less than, or equal to23.2o

? Justify your answer by determining

the angle at which the crate begins to slide (4 points). (b) Find the length of time it

takes for the crate to slide a distance of 2.75mwhen the tilt angle has the values

found in part (a) (4 points).

(iii)The crate begins to slide when the tilt angle is17.5o

. When the crate reaches the

bottom of the flatbed, after sliding a distance of2.75m, its speed is3.11 / m s. Find

(a) the coefficient of static friction (4 points) and (b) the coefficient of kinetic friction

between the crate and the flatbed (4 points).

Explanation / Answer

i)mgsinx=umgcosx

or u=tanx

or u=tan23.2

=0.4286

ii)a)it begins to slide at an angle less than 23.3 since u static < tan23.2

tanx=0.415

or x=22.538 degrees

b)balancing force,

mgsinx-umgcosx=ma

or a=0.2958 m/s^2

so t=(2h/a)^0.5

=4.31 s

a)let the kinetic friction coefficient be x.so,

conserving energy,

0.5*mv^2=mghsink -xmgcosk

or 0.5*3.11^2=9.8*2.75*sin17.5-x*9.8*cos17.5*2.75

or x=0.1264

us=tan17.5

=0.315

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