An accelerating crane system consisting of two masses connected by a cable that passes through a massless pulley is shown below. The mass mj slides in a plane that is inclined by an angle e while the mass m2 moves in air and does no experience air resistance. www WVs旺 m2 If the tension in the cable connecting the two masses is T and mj experiences a kinetic friction whose coefficient is Hk, what is the net force acting on m1? O EF =T+m,g sine + µkm1g cose = m,a O EF=m,g sine – T- Hkm1g cose = m,a O EF =T+mg sin@ – Hym1g cose = m,a O EF =T-m,g – Hxm1g = m¡a O EF=T-m,g sine - Hkm1g cose = m,a O EF =T-m,g sin@ – Hkm1g = m¡a O EF=m1g sin® – T- Hkm1g cose = 0

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Chapter4: The Laws Of Motion
Section: Chapter Questions
Problem 30P: A block of mass m = 5.8 kg is pulled up a = 25 incline as in Figure P4.24 with a force of magnitude...
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An accelerating crane system consisting of two masses connected by a cable that passes through a massless pulley is shown below. The mass m1 slides in a plane that is inclined by an angle e while the mass m2 moves in air and does not
experience air resistance.
m1
m2
If the tension in the cable connecting the two masses is T and m1 experiences a kinetic friction whose coefficient is HK, what is the net force acting on m1?
O F=T+m,g sine + µKm1g cose = m1a
O EF =m1gsine – T- Hkm1g cose = m1a
ο ΣF-T+mjg sinθ
- Hkm1g cos0 = m,a
O EF=T-m1g – Hkm19 = mịa
O F =T-m1g sine – Hkm1g cose = m¡a
O EF=T-m1g sine – Hkm19 = mịa
O EF=m1g sine – T- Hkm1g cose = 0
Transcribed Image Text:An accelerating crane system consisting of two masses connected by a cable that passes through a massless pulley is shown below. The mass m1 slides in a plane that is inclined by an angle e while the mass m2 moves in air and does not experience air resistance. m1 m2 If the tension in the cable connecting the two masses is T and m1 experiences a kinetic friction whose coefficient is HK, what is the net force acting on m1? O F=T+m,g sine + µKm1g cose = m1a O EF =m1gsine – T- Hkm1g cose = m1a ο ΣF-T+mjg sinθ - Hkm1g cos0 = m,a O EF=T-m1g – Hkm19 = mịa O F =T-m1g sine – Hkm1g cose = m¡a O EF=T-m1g sine – Hkm19 = mịa O EF=m1g sine – T- Hkm1g cose = 0
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