Answer: A) 45.5N b) 34.6N
Explanation:
Given the following :
Mass of block A(m1) = 5kg
Mass of block B (m2) = 6kg
Angle of incline (Θ) = 45°
Force pulling the down the incline = m1×a×sinΘ
Tension (T) = (m2×a) - (m2)×g
Where g = acceleration due to gravity
a = acceleration
Net force = (5×9.8×sin45°) - (6×9.8)
Net force = 34.648232 - 58.8
The net force acting on the body = 24.4N
Therefore,
Acceleration, a = Net force/ total mass
a = 24.4 / (6+5)
a = 2.22ms^-2
T = (m2 ×g) - (m2×a)
T = (6 × 9.8) - (6 × 2.22)
T = 58.8 - 13.32
T = 45.48
T = 45.5N
B) Normal reaction:
Horizontal component:
m1gCosΘ = 5 × 9.8 × cos45°
= 5 × 9.8 × 0.7071067
= 34.648232
= 34.6N
The tension on the string is 45.62N and the normal force is 34.65N
Data;
Taking the two mass into consideration
for the 6kg mass
[tex]6g - 6a[/tex]
for the 5kg mass
[tex]5gsin45 + 5a\\[/tex]
Equating the tension in both equations
[tex]6g - 6a = 5gsin45 + 5a[/tex]
Let's solve for a
[tex]6a + 5a = 6g - 5gsin45\\11a = 6g - 5gsin45\\a = \frac{6g - 5gsin45}{11}\\ a = \frac{6* 9.8 - 5*9.8*sin45}{11}\\ a = 2.1956 m/s^2\\T = 6g - 6a\\T = 6*9.8 - 6*2.1956\\T = 45.62N[/tex]
b)
The normal force on the 5kg block.
The normal force on the 5kg block can be calculated as
[tex]N = 5 * 9.8 cos 45 = 34.65N[/tex]
The tension on the string is 45.62N and the normal force is 34.65N
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