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Caerus
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My question is if I actually did the said problem correctly and if there is perhaps an easier way of coming across an answer. Thanks!
There are two blocks present (A and B) A is connected to B via a rope and the applied force is affecting B (to the right).
Find the force of tension in the string if the coefficient of friction is u=0.13
mA = 40 kg
mB = 25 kg
Fapplied = 150 N
u = 0.13
See below.
Tapplied – Tblock – μmBg = mBa
Tblock – μmAg = mAa
Tapplied – μmBg – μmAg = a
mA + mB
Tblock – μmAg = mAa
Tblock – μmAg = mA(Tapplied – μmBg – μmAg)
mA + mB
Tblock = mA(Tapplied – μmBg – μmAg) + mAg
mA + mB
Tblock = 40(150 – (0.13)(25)(-9.8) – (0.13)(40)(-9.8)) + (40)(-9.8)
40 + 25
Tblock = 40(150 + 31.85 + 50.96) + (-392)
65
Tblock = 40(232.81) + (-392)
65
Tblock = 40(3.58) + (-392)
Tblock = 143.3 + (-392)
Tblock = -248.73
Homework Statement
There are two blocks present (A and B) A is connected to B via a rope and the applied force is affecting B (to the right).
Find the force of tension in the string if the coefficient of friction is u=0.13
mA = 40 kg
mB = 25 kg
Fapplied = 150 N
u = 0.13
Homework Equations
See below.
The Attempt at a Solution
Tapplied – Tblock – μmBg = mBa
Tblock – μmAg = mAa
Tapplied – μmBg – μmAg = a
mA + mB
Tblock – μmAg = mAa
Tblock – μmAg = mA(Tapplied – μmBg – μmAg)
mA + mB
Tblock = mA(Tapplied – μmBg – μmAg) + mAg
mA + mB
Tblock = 40(150 – (0.13)(25)(-9.8) – (0.13)(40)(-9.8)) + (40)(-9.8)
40 + 25
Tblock = 40(150 + 31.85 + 50.96) + (-392)
65
Tblock = 40(232.81) + (-392)
65
Tblock = 40(3.58) + (-392)
Tblock = 143.3 + (-392)
Tblock = -248.73