Worksheet 4A Question 1: Determine the tension in each cable in case A and case B.
As shown in the force diagrams, the tension (ceiling on ball) and weight (Earth on ball) forces are equal since the balls are stationary (not moving up or down). In addition, no forces are acting on them from the sides.
To find the tension in the cable, we must find the total force. Our equation to do so is sum of forces equals mass times acceleration [ΣF = m(kg) * a(m/s^2)].
The two forces acting upon the ball are tension (ceiling on ball) and weight (Earth on ball); we combine these two forces to get the sum of forces. Since the tension force pulls the ball upward and the weight force is exerted downward, we make the tension force positive in our equation and the weight force negative. Since the balls are merely hanging from the ceiling (rather than moving up or down), we know that the tension and weight forces must be the same (besides the sign in front of them, which should be opposites). We must find what our weight force is equal to by using the equation weight equals mass times gravity, or W = m(kg) * g(m/s^2). We then plug in the mass (5kg) and gravity (9.8m/s^2). On the right side of our total force equation, we plug in mass (5kg), but 0m/s^2 for acceleration since there is no motion.
Once we solve, we find that the tension force is equal to 49N.
To find the tension in Case B, we must include 2 tension forces in the equation since there are 2 cables pulling on the ball. We then follow the same steps as we did in the math for A.
To find the tension in Case B, we must include 2 tension forces in the equation since there are 2 cables pulling on the ball. We then follow the same steps as we did in the math for A.
We now know that the tension for each cable in Case B is equal to 24.5N.
Worksheet 4 Question 2: An elevator is accelerating upward at 2.0m/s^2. The man has a mass of 85.kg.
a. Construct a force diagram for the man
b. What force does the floor now exert on the man?
a. The normal force is larger than the weight force since the man accelerates upward. There are no forces acting upon the man sideways.
b. We are trying to find the support force (also known as the normal force) that the floor of the elevator has on the man. Again, we use the equation sum of forces equals mass times acceleration to find the force. This time, only weight and normal forces are acting on the man. The normal force is causing the man to accelerate upward, so we use +N in our equation. The weight force is pulling him down towards Earth, so it is signified as -W. We substitute mass times gravity for W (weight) on the left side of the equation since we know the man's mass and gravity.
Once we solve, we find that the normal support force is 1003N.
Worksheet 6 Question 2: A block weighing 300.N is moved at constant speed over a horizontal surface by a force of 50.N applied parallel to the surface.
a. Construct a force diagram for the block.
The problem tells us that the weight is 300.N, and since the block is moving horizontally (rather than vertically), we know that the normal and weight forces must be equal. Therefore, the normal force must also equal 300.N. We are also given that the force applied horizontally is 50.N. We are choosing to include a friction force opposing the motion.
What is the coefficient of kinetic friction?
To find the coefficient of kinetic friction, we use the formula kinetic friction equals coefficient of friction times normal force (fk = µk * N). In this case, we are given the force which is being applied (50.N). This is also equal to the amount of kinetic friction, so we plug it in for the fk in the formula. We are also given the weight of the block. Because the force is moving the block horizontally on a surface, we know that the block is mnot moving up or down. Therefore, the normal and weight forces on the block must be equal so they will cancel each other out (as shown in the force diagram above). This means that 300.N can be substituted for N in our equation.
When we solve, we get that µk = .167.
Worksheet 4 Question 2: An elevator is accelerating upward at 2.0m/s^2. The man has a mass of 85.kg.
a. Construct a force diagram for the man
b. What force does the floor now exert on the man?
a. The normal force is larger than the weight force since the man accelerates upward. There are no forces acting upon the man sideways.
b. We are trying to find the support force (also known as the normal force) that the floor of the elevator has on the man. Again, we use the equation sum of forces equals mass times acceleration to find the force. This time, only weight and normal forces are acting on the man. The normal force is causing the man to accelerate upward, so we use +N in our equation. The weight force is pulling him down towards Earth, so it is signified as -W. We substitute mass times gravity for W (weight) on the left side of the equation since we know the man's mass and gravity.
ΣF = m(kg) * a(m/s^2)
+N -W = 85.kg * 2.0m/s^2
N - (m * g) = 170N
N - (85.kg * 9.8m/s^2) = 170N
N - 833N = 170N
N = 1003N
Once we solve, we find that the normal support force is 1003N.
Worksheet 6 Question 2: A block weighing 300.N is moved at constant speed over a horizontal surface by a force of 50.N applied parallel to the surface.
a. Construct a force diagram for the block.
The problem tells us that the weight is 300.N, and since the block is moving horizontally (rather than vertically), we know that the normal and weight forces must be equal. Therefore, the normal force must also equal 300.N. We are also given that the force applied horizontally is 50.N. We are choosing to include a friction force opposing the motion.
What is the coefficient of kinetic friction?
To find the coefficient of kinetic friction, we use the formula kinetic friction equals coefficient of friction times normal force (fk = µk * N). In this case, we are given the force which is being applied (50.N). This is also equal to the amount of kinetic friction, so we plug it in for the fk in the formula. We are also given the weight of the block. Because the force is moving the block horizontally on a surface, we know that the block is mnot moving up or down. Therefore, the normal and weight forces on the block must be equal so they will cancel each other out (as shown in the force diagram above). This means that 300.N can be substituted for N in our equation.
fk = µk * N
50.N = µk * 300.N
µk = 50.N/300.N = .167
When we solve, we get that µk = .167.
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