Answer:
a= 0.578 m/s
T = 1037.8 N
Explanation:
Data
m₁= 300 kg
m₂= 100 kg
inclined plane, θ = 30°
μk = 0.120
Newton's second law to m₁:
We define the x-axis in the direction parallel to the movement of the 300kg (m₁) crate on the ramp and the y-axis in the direction perpendicular to it.
∑F = m₁*a Formula (1)
Forces acting on m₁
W₁: m₁ weight : In vertical direction
N : Normal force : perpendicular to the inclined plane
f : Friction force: parallel to the inclined plane
T: cable tension : parallel to inclined plane
Calculated of the W₁
W₁=m₁*g
W₁= 300kg* 9.8 m/s² = 2940 N
x-y weight components
W₁x= W₁sin θ =2940 N*sin(30)° =1470 N
W₁y= W₁cos θ =2940 N *cos(30)° =2156.4 N
Calculated of the N
We apply the formula (1)
∑Fy = m*ay ay = 0
N - W₁y = 0
N = W₁y
N = 2156.4 N
Calculated of the f
f = μk* N= (0.120)*(2156.4 N)
f = 258.77 N
Newton's second law to m₁ in direction x-axis :
∑Fx = m₁*ax ,ax =a
We assume that m₁ descends on the inclined plane and we positively take the direction of movement:
wx-f-T = m*a
wx - f - m*a =T
1470 -258.77 -300*a =T
T= 1211.23-300*a Equation (1)
Newton's second law to m₂
∑Fy = m₂*ay ,ay =a
Forces acting on m₂
W₂: m₂ weight : In vertical direction
T: cable tension:In vertical direction
Calculated of the W₂
W₂=m₂*g
W₂= 100kg* 9.8 m/s² = 980 N
∑Fy = m₂*a
Because we assume that m₁ descends on the inclined plane, then, m₂ ascends vertically, we take positive the direction of movement:
T-W₂ = m₂*a
T-980 = 100*a
T = 980 + 100*a Equation (2)
Problem development
Equation (1) = Equation (2) = T
1211.23-300*a= 980 + 100*a
1211.23- 980 = 100*a + 300*a
231.23 = 400*a
a= 231.23 / 400
a= 0.578 m/s
Because the acceleration tested positive then effectively m₁ descends on the inclined plane and m₂ ascends vertically.
We replace a= 0.578 m/s in the equatión (2)
T = 980 + 100* (0.578 )
T = 1037.8 N
