Answer:
T1 = 130N, T2 = 370N
Explanation:
In order for the system to be at rest, the sum of all forces must be zero and the torque around a point on the beam must be zero.
1. forces:
Let tension in rope 1 be T1 and in rope 2 be T2:
ma = T1 + T2 - 100N - 400N = 0
(1) T1 + T2 = 500N
2. torque around the center point of the beam:
τ = r x F = 5*T1 + 3*400N - 5*T2 = 0
(2) T1 - T2 = -240N
Solving both equations:
T1 = 130N
T2 = 370N
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Answer:
3.2N
Explanation:
Given parameters:
Mass of block = 1.5kg
Coefficient of kinetic friction = 0.6
Force of pull on block = 12N
Unknown:
Net force on the block = ?
Solution:
Frictional force is a force that opposes motion:
Net force = Force of pull - Frictional force
Frictional force = umg
u is coefficient of kinetic friction
m is the mass
g is the acceleration due to gravity
Frictional force = 0.6 x 1.5 x 9.8 = 8.8N
Net force = 12N - 8.8N = 3.2N
