An inclined plane making an angle of 25 degrees with the horizontal has a pulley at its top. A 30 kg block on the plane is connected to a freely hanging 20 kg block by means of a cord passing over the pulley. Compute the distance the 20 kg block will fall in 2.0 seconds starting from rest. The coefficient of friction between block and plane is 0.20.
To start the problem, I decided to draw out the scenario to understand all the forces acting on the objects. The diagram is below.
Let A be the 30 kg weight and B be the 20 kg weight. We can now proceed to draw the force diagrams for both A and B. For A, as gravity is pulling down on it, there is a weight force directly downward. This is symbolized in the picture as Wea (Weight from earth onto block A). The support force is also present as the ramp is pushing on A. This force is perpendicular to the ramp surface and is notated Nra (the normal force of the ramp onto block A). B also exerts a force on A, this is a tension force that is transmitted through the rope connecting the blocks. This is also at an angle because the string is parallel to the ramp surface. This is symbolized by Tba (Tension exerted by block B on block A). Finally, there is a frictional force resisting motion; this is also at the same angle as the tension force and is notated with f (for friction). We can now draw a free-body diagram for A. It is pictured below.
To make it so that only 1 force has to be broken into components, the diagram can be rotated slightly and the weight force can be broken into components. Because the object isn't accelerating in the vertical direction, we can see that the normal force (Nra) and the y component of the weight force (Weay) are balanced. We also know that the forces in the horizontal direction aren't balanced because the block is accelerating.
Next, we consider the forces on object B. There is a downward weight force due to earth's gravitational pull (symbolized by Web), but there is no normal force to counteract it. Instead, the tension force transmitted through the rope as A pulls on B (symbolized with Tab). Since there is acceleration in the downward direction, we know that the weight force is larger than any force counteracting it. There are no forces affecting B in the horizontal direction. The free-body diagram is pictured below:
Next, we consider the forces on object B. There is a downward weight force due to earth's gravitational pull (symbolized by Web), but there is no normal force to counteract it. Instead, the tension force transmitted through the rope as A pulls on B (symbolized with Tab). Since there is acceleration in the downward direction, we know that the weight force is larger than any force counteracting it. There are no forces affecting B in the horizontal direction. The free-body diagram is pictured below:
It should also be mentioned that Tab and Tba are equal and opposite forces, according to Newton's 3rd Law. To solve for the distance B will fall, we must know its acceleration. To find its acceleration, we must know the acceleration of the entire system because the acceleration of individual elements in a system equals that of the entire system. The net force acting on the system can be attributed to 3 forces: Web, Wea(x), and f. These three forces are the only forces that impact the outside of the system, and are thus the only forces able to cause acceleration. We can use these 3 forces in Newton's 2nd Law.
ΣF = m*a
Web - f - Wea (x) = m*a
Web - f - Wea (x) = m*a
To find Web, we can use another net force equation and plug in acceleration due to gravity to help us find it.
ΣF = m*a
Web = 20 kg * 9.8 m/s^2
Web = 196N
Web = 20 kg * 9.8 m/s^2
Web = 196N
Similarly, we can find Wea and use trigonometric ratios to solve for Wea(x).
ΣF = m*a
Wea = 30 kg * 9.8 m/sec^2
Wea = 294N
sin25 = Wea(x)/Wea
294sin25 = Wea(x)
Wea(x) = 124.5N
Finally, to calculate friction, we must first find the normal force on object A. We can set up yet another net force equation in the vertical direction to calculate the normal force, Nra.
ΣF = m*a
Nra - Wea(y) = m*a
Nra - Wea(y) = 30 kg * 0
To calculate Wea(y), we can use a similar strategy to what we did to find Wea(x), but we must use cosine instead of sine to get the correct ratio of sides.
cos25 = Wea(y)/Wea
294cos25 = Wea(y)
Wea(y) = 266.45N
We can insert this value back into the previous net force equation to solve for the normal force.
Nra - 266.45 = 0
Nra = 266.45N
Now we can calculate friction because the friction and normal forces are directly proportional, and the problem gave us the frictional constant.
f = μN
f = 0.2 * 266.45N
f = 53.29N
Now that we have calculated all of the forces that impact the entire system, we can take these values and plug them into the net force equation for the entire system. Because the system contains a total of 50 kg, we must use that number in the equation.
Web - f - Wea (x) = m*a
196N - 53.29N - 124.25N = 50*a
a = 0.3692m/sec^2
Using this acceleration, we can now construct a quantitative vt graph for the movement of the system. Acceleration is the slope of the velocity. Because the object is moving downward, we will use the negative of the acceleration to calculate displacement.
Wea = 30 kg * 9.8 m/sec^2
Wea = 294N
sin25 = Wea(x)/Wea
294sin25 = Wea(x)
Wea(x) = 124.5N
Finally, to calculate friction, we must first find the normal force on object A. We can set up yet another net force equation in the vertical direction to calculate the normal force, Nra.
ΣF = m*a
Nra - Wea(y) = m*a
Nra - Wea(y) = 30 kg * 0
To calculate Wea(y), we can use a similar strategy to what we did to find Wea(x), but we must use cosine instead of sine to get the correct ratio of sides.
cos25 = Wea(y)/Wea
294cos25 = Wea(y)
Wea(y) = 266.45N
We can insert this value back into the previous net force equation to solve for the normal force.
Nra - 266.45 = 0
Nra = 266.45N
Now we can calculate friction because the friction and normal forces are directly proportional, and the problem gave us the frictional constant.
f = μN
f = 0.2 * 266.45N
f = 53.29N
Now that we have calculated all of the forces that impact the entire system, we can take these values and plug them into the net force equation for the entire system. Because the system contains a total of 50 kg, we must use that number in the equation.
Web - f - Wea (x) = m*a
196N - 53.29N - 124.25N = 50*a
a = 0.3692m/sec^2
Using this acceleration, we can now construct a quantitative vt graph for the movement of the system. Acceleration is the slope of the velocity. Because the object is moving downward, we will use the negative of the acceleration to calculate displacement.
They have specified to calculate displacement at t= 2.0 seconds. To do this, we must first calculate the equation of the velocity line. Because this describes velocity, we can use the general equation V= at + Vo, where a= -0.3692 and Vo= 0 (this is because the system started at rest). Hence, the equation of the line is
v = -0.3692t. Since displacement equals the area under the velocity curve, and the area depicted above is a triangle, we can use the general equation for a triangle.
v = -0.3692t. Since displacement equals the area under the velocity curve, and the area depicted above is a triangle, we can use the general equation for a triangle.
A = (1/2) b*h
Δx = (1/2) (2) (-0.3692(2))
Δx = -0.7384 meters
Δx = (1/2) (2) (-0.3692(2))
Δx = -0.7384 meters
The block will fall 0.7384 meters in 2 seconds after being released from a position of rest.