A block of mass m1 hangs vertically and is attached by a string to a block of mass m2 on a frictionless slope inclined at angle theta to the horizontal, as shown above. If m1 =
2.5 kg, it accelerates at -0.108 m/s2. If m1 = 0.8 kg, it accelerates at 2.888 m/s2.
Determine the angle theta and m2.
m2 going up is +
-m1 g + m2 g sin T= (m1+m2)a
-2.5(9.8)+9.8m2 sin T=-(2.5+m2).108
-.8(9.8)+9.8m2sinT = (.8+m2)2.888
----------------------------------
subtract to get rid of 9.8m2 sin T
solve result for m2
go back and get m2 sinT and therefore sin T
To solve this problem, we can use Newton's second law of motion and apply it separately to each block.
Let's consider the block with mass m1 first. The forces acting on this block are the tension in the string (T) and the force due to gravity (mg).
The equation of motion for m1 can be written as:
m1 * (-0.108 m/s^2) = T - m1 * g (Equation 1)
Next, let's consider the block with mass m2. The forces acting on this block are the force due to gravity (m2 * g) and the component of the tension along the inclined plane (T * sin(theta)).
Since the block is on a frictionless slope, the acceleration along the slope is given by a = g * sin(theta). Therefore, we can write the equation of motion for m2 as:
m2 * a = m2 * g * sin(theta) - T * sin(theta) (Equation 2)
Now we have two equations (Equation 1 and Equation 2) with two unknowns (T and theta). We can solve this system of equations to find the values of T and theta.
To find the angle theta, we can divide Equation 2 by m2:
a = g * sin(theta) - (T / m2) * sin(theta)
Rearranging the equation:
(T / m2) * sin(theta) = g * sin(theta) - a
Dividing both sides by sin(theta):
(T / m2) = g - (a / sin(theta))
Now we have an equation with only one unknown, theta. We can substitute the known values of m1, m2, and a, and solve for theta.
Once we have the value of theta, we can substitute it back into Equation 1 to solve for T.
Finally, we can use the value of T to find m2 by substituting it back into Equation 2 and solving for m2.