tem 10 Part A After the system is released, find the horizontal tension in the w

Question

tem 10

Part A

After the system is released, find the horizontal tension in the wire.

Incorrect; Try Again; 6 attempts remaining

Part B

After the system is released, find the vertical tension in the wire.

Part C

After the system is released, find the acceleration of the box.

Part D

After the system is released, find magnitude of the horizontal and vertical components of the force that the axle exerts on the pulley.

Express your answers separated by a comma.

tem 10

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 1) ). The pulley has the shape of a uniform solid disk of mass 1.50kgand diameter 0.580m .

Part A

After the system is released, find the horizontal tension in the wire.

|Th| = ??   N  

Incorrect; Try Again; 6 attempts remaining

Part B

After the system is released, find the vertical tension in the wire.

|Tv| = ??   N  

Part C

After the system is released, find the acceleration of the box.

a = ??   m/s2  

Part D

After the system is released, find magnitude of the horizontal and vertical components of the force that the axle exerts on the pulley.

Express your answers separated by a comma.

Fx,Fy = ??   N

Explanation / Answer

mass of pulley ( solid disk) =1.50kg

radius= diameter/2 = 0.480 / 2 =0.24 m

moment of inertia=(1/2mr^2) = 0.5*1.5*(0.24)^2=0.0432 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/0.24

Net torque=[ Tv -Th]*r=[Tv-Th ]0.24

Net torque=[49 - 5a- 12 a ]0.24

Net torque=[49-17a]0.24

but net torque= I alpha= Ia /r=0.0432 a/0.24=0.18 a

0.18a=[49-17a]0.24-----------(3)

a=2.7606 m/s^2

Th=12a=33.126 N

(A) horizontal tension is 33.126 N

(B)vertical tension=Tv=49 -5a=35.197 N

vertical tension is 35.197 N
(C)acceleration of box is 2.7607 m/s^2

(D)magnitude of the horizontal component of the force F that the axle exerts on the pulley=Th =33.126 N

magnitude of the vertical components of the force that the axle exerts on the pulley= Tv + weight of wheel= 35.197+ 14.7=49.267 N

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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