A block of mass(m) is being pulled at a constant speed(v) down a slope that makes an angle(theta) with the horizontal. The pulling force is applied through a horizontal rope, as shown below. If the coefficient of kinetic friction is Uk, find an expression for the rope tension.
2006-11-16 11:06:02 · 3 answers · asked by Thomas V 1 in Science & Mathematics ➔ Physics
(Fn)(Uk) = Fk
Fk = Fp
ANSWER
(Fn)(Uk) = Fp
2006-11-16 11:11:59 · answer #1 · answered by trackstarr59 3 · 0⤊ 0⤋
The constant speed says that the friction is just being met by influences to move the block down the slope. The friction is
Ff = Uk*N = Uk*m*g*cos(theta).
A component of the weight parallel with the slope, Wp, is helping the block move down the slope.
Wp = m*g*sin(theta)
A component of the rope tension parallel with the slope, Tp, is helping it move down the slope.
Tp = T*cos(theta)
Ff = Wp + Tp
Substitute and simplify.
2006-11-16 12:15:34 · answer #2 · answered by sojsail 7 · 0⤊ 0⤋
i don't love numbers, so i will replace the equipment weight with W, and the given perspective with theta. rigidity in higher cable is T1 rigidity in decrease cable is T2 The device isn't accelerating, so all forces could upload as a lot as 0. Vertical rigidity stability: T1*sin(theta) = W Horizontal rigidity stability: T1*cos(theta) = T2 for this reason: T1 = W/sin(theta) T2 = T1*cos(theta)/sin(theta) = W/tan(theta) documents: W:=500 lbs; theta:=30 deg; outcomes: T1 = one thousand lbs T2 = 866 lbs
2016-11-24 23:19:00 · answer #3 · answered by ? 4 · 0⤊ 0⤋