1.) A stubborn 140 kg pig sits down and refuses to move. To drag the pig to the

Question

1.) A stubborn 140 kg pig sits down and refuses to move. To drag the pig to the barn, the exasperated farmer ties a rope around the pig and pulls with his maximum force of 800 N. The coefficients of friction between the pig and the ground are ?s=0.80 and ?k=0.50. Part A Calculate the force which farmer needs to apply to budge the pig. Express your answer to two significant figures and include the appropriate units.

Static Friction and Frictional Force Ranking Task Below are six crates at rest on level surfaces. The crates have different masses and the frictional coefficients [given as (mu s, mu k)] between the crates and the surfaces differ. The same external force is applied to each crate, but none of the crates move. Part A Rank the crates on the basis of the frictional force acting on them. Rank from largest to smallest. To rank items as equivalent, overlap them. Problem 5.65 A person with compromised pinch strength in their fingers can only exert a normal force of 6.0 N to either side of a pinch-held object such as the book shown in (Figure 1).

Explanation / Answer

1)

until unless the pig moves static friction acts on it once the pig starts moving kinetic friction comes in to play

so when the pig is not moving=frictional force acting on it =normal force*co-efficeint of static friction

=140*9.8*0.8=1097.6 newtons

so farmer should apply more than 1097.6 N intially to start

once it is started to move than kinetic friction acts so

140*9.8*0.5=686 newtons

so once when it is started to move he can apply more than 686 newtons to pull it

2)

the crates are not moving so static friction is acting on them

so friction force=m*9.8*co-efficient of static friction

for 1st crate
4410

2) 490
3) 4704
4) 2940
5) 2940
6) 4410

3>6=1>4=5>2

3)

max tension will be at top and min at bottom

at top tension =

mass of loud speaker*g+mass of cable*g=27*9.8/2+60*9.8=1884.54 newtons (in one cable)

at bottom only due to loud speaker=27*9.8/2=132.3 newtons (in one cable)

at middle=mass of cable*g/2+mass of loud speaker*g/2=60*9.8/2+27*9.8/2=294+132.3=426.3 newtons (in one

4)

friction force=

6*0.65=3.9 newtons

so the highest weight of the book = 3.9/9.8=0.397 kg

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