Physics 140
HOMEWORK Chapter 5B
~.
Q10. Figure 5-27 shows three blocks being pushed across a frictionless floor by horizontal force F
~ , (b) force F
~ 21 on block 2 from block 1, and
What total mass is accelerated to the right by (a) force F
~
(c) force F32 on block 3 from block 2? (d) Rank the blocks according to their acceleration magnitudes,
~, F
~ 21 , and F
~ 32 , according to magnitude, greatest first.
greatest first. (e) Rank forces F
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(a) 17 kg. One way to look at it is that all three blocks are effectively one unit.
(b) 12 kg. Similarly
(c) 10 kg: only the last block.
(d) All three accelerations are equal
~ >F
~ 21 > F
~ 32 .
(e) F
P33. An elevator cab and its load have a combined mass of 1600 kg. Find the tension in the supporting
cable when the cab, originally moving downward at 12 m/s, is brought to rest with constant acceleration
in a distance of 42 m.
——–
~ NET = m~a, and its x and y components, are understood throughout
Newton’s 2nd Law F
as the starting point of equations. I will substitute into them the forces on the LHS and
the ma on the RHS. Also, I describe, but don’t draw, FBDs. You need to draw them.
FBD: T up; mg down. ~a upward (!). Take +x upward.
T − mg = ma ⇒ T = m(g + a). All we need is a, but that’s just 1-D kinematics:
v 2 − v02 = 2a∆x ⇒ a = (v 2 − v02 )/2∆x = (0 − (−12)2 )/(2 · (−42 m)) = 1.714 m/s2 .
T = (1600 kg)(9.8 m/s2 + 1.714 m/s2 ) = 18.4 kN.
P39. A sphere of mass 3.0 × 10−4 kg is suspended from a cord. A steady horizontal breeze pushes the
sphere so that the cord makes a constant angle of 37◦ with the vertical. Find (a) the push magnitude
and (b) the tension in the cord.
———
~ to right. ~a = 0. Make coordinate system x to
FBD: mg down; T up/left at 37◦ from vertical (!); P
right, y upward.
x-eq: P − T sin θ = 0.
y-eq: T cos θ − mg = 0.
We note the x-eq has 2 unknowns (T and P ), while the y-eq has only one. Solve the y-eq for T
T = mg/ cos θ = (3 × 104 )(9.8)/0.7986 = 3.68 × 10−3 N.
Solve the x-eq for P , since T is now known.
(a) P = T sin θ = (3.68 × 10−3 N)(sin 37◦ ) = 2.22 × 10−3 N.
(b) We got that above: T = 3.68 × 10−3 N.
P50. In Fig. 5-46, three ballot boxes are connected by cords, one of which wraps over a pulley having
negligible friction on its axle and negligible mass. The three masses are mA = 30.0 kg, mB = 40.0 kg,
and mC = 100.0 kg, When the assembly is released from rest, (a) what is the tension in the cord
connecting B and C, and (b) how far does A move in the in the first 0.250 s (assuming it does not
reach the pulley)?
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You should realize that the tension in the lower cord cannot be calculated without the acceleration,
and that an analysis of the full system is needed to calculate the acceleration. So: there isn’t much of
a shortcut. To get the acceleration, we can consider B and C to be one object.
FBD for A: T1 to right; n upward; mA g downward. ~a to right. Make +x rightward. T1 is the tension
in the cord that goes over the pulley.
FBD for BC: (mB + mC )g downward; T1 upward. ~a downward. Make +x downward (!). The
acceleration of all three blocks is equal, positive, and in the +x direction.
Maybe you can already see that the y-eq for A gives us that n = mA g, and that this has no bearing on
the acceleration in a frictionless problem.
x-eq for A: T1 = mA a
x-eq for BC: (mB + mC )g − T1 = (mB + mC )a. Add these two equations:
mB + mC )g = (mA + mB + mC )a ⇒
a = g(mB + mC )/(mA + mB + mC ) = (9.8)(50/80) = 6.125 m/s2 .
(a) Now the FBD for C: T2 up; mC g down. ~a down. Make +x downward.
x-eq: mC g − T2 = mC a ⇒ T2 = mC (g − a) = (10 kg)(9.8 m/s2 − 6.125 m/s2 ) = 36.7 N. (Note
that while you are holding the system, T2 = 98 N.)
(b) ∆x = v0 t + (1/2)at2 = 0 + (1/2)(6.125(0.250)2 = 0.191 m.
P57. A block of mass m1 = 3.70 kg on a frictionless plane inclined at angle θ = 30.0◦ is connected by a
cord over a massless, frictionless pulley to a second block of mass m2 = 2.30 kg (Fig. 5-52). What are
(a) the magnitude of the acceleration of each block, (b) the direction of the acceleration of the hanging
block, and (c) the tension in the cord?
———
(a) FBD for m1 : m1 g down; T up/right; n away from plane. ~a downplane (guessed). +x downplane.
FBD for m2 : T up; m2 g down. ~a upward (to be consistent with the guess for part (a). +x upward.
With these choices for coordinate systems, a1y = 0, and both blocks have the same ax , which I will call
ax .
y-eq for m1 : n − m1 g cos θ = 0 ⇒ n = m1 g cos θ.
x-eq for m1 : m1 g sin θ − T = m1 ax .
x-eq for m2 : T − m2 g = m2 ax . Add the two x-eqs to eliminate T :
(m1 sin θ − m2 )g = (m1 + m2 )ax ⇒
ax = (m1 sin θ − m2 )g/(m1 + m2 ) = −0.735 m/s2 . So
|~a| = 0.735 m/s2 .
(b) Downward. We guessed upward and got a negative answer. In a frictionless problem, this just
means we guessed wrong on the direction. With a friction problem, we would have to reverse the
direction of the frictional force f and re-evaluate. There would always be the possibility that it doesn’t
move at all.
(c) From the x-eq for m2 ,
T = m2 (g + ax ) = (2.3 kg)(9.8 m/s2 − 0.735 m/s2 ) = 20.8 N. Keep in mind that ax was negative,
and that this tension is less than the gravitational force of m2 g = 22.5 N.
P67. Figure 5-58 shows three blocks attached by cords that loop over frictionless pulleys. Block B
lies on a frictionless table; the masses are mA = 6.00 kg, mB = 8.00 kg, and mC = 10.0 kg. When the
blocks are released, what is the tension in the cord at the right?
———
Let T1 be the tension in the left cord and T2 the tension in the right cord. We are asked for only T2 ,
but by now realize that we only get that by essentially solving for T1 and ax . Since mC > mA , it is
clear that B moves to the right .
FBD for A: T1 up, m1 g down. ~a upward. +x upward.
FBD for B: T1 to left; T2 to right; n up; mB g down. ~a to right. +x to right.
FBD for C: T2 up; mC g down. ~a downward. +x downward.
These choices of coordinate systems mean that ax is equal for all three blocks, and is positive.
y-eq for B: n − mB g = 0 ⇒ n = mB g. This isn’t involved in the tensions in a frictionless problem.
x-eq for A: T1 − mA g = mA ax .
x-eq for B: T2 − T1 = mB ax .
x-eq for C: mC g − T2 = mC ax .
The easiest way (I think) is to add all three equations, get ax , then use the x-eq for C to get T2 . Adding
the three eqs, T1 and T2 cancel, leaving:
(mC − mA )g = (mA + mB + mC )ax . Then
ax = (mC − mA )g/(mA + mB + mC ) = 1.63 m/s2 .
T2 = mC (g − ax ) = 81.7 N.
Additional exercise: what is T1 , and how does this compare to mA g?
85. (Opt) A 52 kg circus performer is to slide down a rope that will break if the tension exceeds
425 N. (a) What happens if the performer hangs stationary on the rope? (b) At what magnitude of
acceleration does the performer just avoid breaking the rope?
———
(a) Assume the person is stationary. Calculate tension then compare. FBD for person: mg down; T
up. ~a = 0. Make +x upward (not critical).
x-eq: T − mg = ma − x = 0 ⇒ T = mg = (52 kg)(9.8 m/s2 ) = 509.6 N. Since that exceeds the limit
of 425 N, the rope breaks.
(b) FBD is similar. We might replace T by fK since the performer is sliding down the rope. But fK at
the performer is balanced by T in the rope, so thety are equal. The other change is that ~a is definitely
downward. This tiome, I will leave +x upward, so that we expect ax to be negative.
x-eq: fK − mg = max ⇒ ax = (fK − mg)/m = −1.63 m/s2 .
This means the acceleration is 1.63 m/s2 downward.