A 12.0-kg box resting on a horizontal, frictionless surface is attached to a 5.0

Question

A 12.0-kg box resting on a horizontal, frictionless surface is attached to a 5.00-kg weight by a thin, light wire that passes without slippage over a frictionless pulley (the figure (Figure 11). The pulley has the shape of a uniform solid disk of mass 1.70 kg and diameter 0.460 m. Part A After the system is released, find the horizontal tension in the wire. Part B After the system is released, find the vertical tension in the wire. PartC After the system is released, find the acceleration of the box. My Answers Give Up

Explanation / Answer

________________________________________...
mass of pulley ( solid disk) =1.70kg

radius= diameter/2 = 0.460 / 2 =0.23 m

moment of inertia=(1/2mr^2) = 0.5*1.7*(0.23)^2=0.0449 kgm^2

mass of box = M = 12 kg

acceleration of box= a

net force on box=ma

but net force on box =tension in horizontal portion of wire =Th

Th= 12 a..........................(1)

tension in vertical portion of wire = Tv

weight suspended =mg=5*9.8 =49 N

net force on suspended weight = 49 - Tv

but net force on suspended weight =ma=5a

49 - Tv=5a

Tv=49 -5a .........................(2)

If alpha is angular acceleration of pulley,

alpha=linear acceleration/ radius =a/0.23

Net torque=[ Tv -Th]*r=[Tv-Th ]0.23

Net torque=[49 - 5a- 12 a ]0.23

Net torque=[49-17a]0.23

but net torque= I alpha= Ia /r=0.0449 a/0.23=0.19 a

0.19a=[49-17a]0.23-----------(3)

a=2.748 m/s^2

Th=12a=32.97 N

(A) horizontal tension is 32.97 N

(B)vertical tension=Tv=49 -5a=35.26 N

vertical tension is 35.26 N

(C)acceleration of box is 2.748 m/s^2

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.