Two packing crates of masses m1 = 10.0 kg and m2 = 6.00 kg are connected by a light string that passes over a frictionless pulley as in Figure P4.26. The 6.00 kg crate lies on a smooth incline of angle 42.0°. Find the acceleration of the 6.00 kg crate.
m/s2 (up the incline)
Find the tension in the string.
N
2007-09-25 08:22:38 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Physics
Couldn't find fig 4.26, assuming m1 is on the high level and m2 is on the incline v0 = 0
1. force down is 10 * 6 sin42 = 40N, these 40N need to accelerate 16Kg F = ma => a = F / m = 40 / 16 = 2.5 m/s^2
2. Tension is 40N pulling down the while F = ma = 6*2.5 = 15N absorbs in the m2 acceleration so 40 - 15 = 25N is the tension on the string
2007-09-25 09:21:24 · answer #1 · answered by eyal b 4 · 0⤊ 0⤋
I will assume that mass 1 is hanging and that there is no friction in the system.
The hanging mass has a FBD where
m*g-T=m*a in the direction of motion
m-10
g=9.81
m2 has FBD of
T-m*g*sin(42)=m*a in the direction of motion
because the masses are connected by a string over a massless and frictionless pulley, then
(T-6*9.81*sin(42))/
(10*9.81-T)=10/6
solve for T
T-39.385=(5/3)*(98.1-T)
T=((5*98.1/3)+39.385)/
(1+5/3)
T=76.1 N
j
2007-09-25 16:17:00 · answer #2 · answered by odu83 7 · 0⤊ 0⤋