A mass of 20 kg on a plane inclined at 40 degrees. A string attached to that mass goes up the plane, passed over a pullley and is attached to mass of 30 kg that hangs verticalyy. a) find the acceleration and it's dirction b) the tension in the string. Assume no friction.
I first drew the picture. How would I find the equation since F=ma doesn't include everything?
F=ma (combined with Newton's third law) give you all you need to solve this problem.
The force on the hanging mass in the downward direction is:
F2 = m2 * g - T
where m2 = 30 kg
The force on the other mass in the direction parallel to the plane in which the string is pulling is:
F1 = -m1 * g sin(40°) + T
where m1 = 20 kg
Begause the string is assumed to be of fixed length the acceleration of mass 1 in the direction the string is pulling must be the same as the acceleration of mass 2 in the downward direction.
This means that:
F1/m1 = F2/m2
To find the acceleration (a) and its direction, we can set up a force equation using the equations mentioned above:
F1/m1 = F2/m2
Substituting the expressions for F1 and F2:
(-m1 * g sin(40°) + T)/m1 = (m2 * g - T)/m2
Simplifying and rearranging the equation:
-m1 * g sin(40°) + T = (m2 * g - T) * (m1/m2)
Expanding the right side of the equation:
-m1 * g sin(40°) + T = (m1 * g * (m2/m2)) - (T * (m1/m2))
Simplifying further:
-m1 * g sin(40°) + T = m1 * g - T * (m1/m2)
Rearranging and combining like terms:
2T = m1 * g - m1 * g sin(40°)
2T = m1 * g (1 - sin(40°))
T = (m1 * g (1 -sin(40°)))/2
Using the given values, m1 = 20 kg and g is the acceleration due to gravity (9.8 m/s^2):
T = (20 kg * 9.8 m/s^2 * (1 - sin(40°)))/2
Now, we can calculate T to find the tension in the string.