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Figure 5-27 shows three blocks being pushed across a frictionless floor by horizontal forceF→. What total mass is accelerated to the right by (a) forceF→, (b) forceF→21on block 2 from block 1, and (c) forceF→32on block 3 from block 2? (d) Rank the blocks according to their acceleration magnitudes, greatest first. (e) Rank forcesF→,F→21andF→32, and according to magnitude, greatest first.

Short Answer

Expert verified

a) Total mass accelerated byF→=17kg

b) Total mass accelerated byF→21=12kg

c) Total mass accelerated byF→32=10kg

d) Acceleration will be the same for each situation

e) Rank F→,F→21andF→32according to magnitude (Greatest First)F→>F→21>F→32

Step by step solution

01

Given information

m1=5kgm2=2kgm3=10kg

02

To understand the concept

The problem is based on Newton’s second law of motion which states that the acceleration of an object is dependent on the net force acting upon the object and the mass of the object. Using a free body diagram, how much mass the respective force is accelerating can be calculated. Using Newton's 2nd law of motion, the block’s acceleration and the forces can be ranked.

Formula:

Fnet=ma

03

(a) To find total mass accelerated by F→

F→is pushing all the three blocks, so we conclude that,

role="math" localid="1657009110359" TotalmassacceleratedbyF→is=m1+m2+m3=5kg+2kg+10kg=17kg

04

(b) To find total mass accelerated by F→21

F→21is pushing all the three blocks, so we conclude that,

role="math" localid="1657009265260" TotalmassacceleratedbyF21→=m2+m3=2kg+10kg=12kg

05

(c) To find total mass accelerated by F→32

As we can see in the diagram,

F→32is pushing all the three blocks, so we conclude that,

TotalmassacceleratedbyF32→=m3=10kg

06

(d) To rank the blocks according to their acceleration magnitude (Greatest First)

The value for the acceleration for all three situations will be the same.

07

(e) To Rank F→,F→21 and F→32 according to magnitude (Greatest First)

As the acceleration is the same for all three situations, the magnitude of the force will be based on the mass it is accelerating.

So with this newton’s 2nd law for each force will be,

F→=maF→=17aF21→=maF21→=12aF32→=maF32→=10a

This gives us the relation,

F→>F→21>F32

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