I am really nervous about mathematics, so I just want to make sure I am doing these problems correctly. I figure if I am getting the incorrect answer, I must be doing them incorrectly and need to work on them some more. I THINK I have them correct, but I just want to verify. Even one or two will greatly help, if you cannot verify all of them. The images that apply to these questions can be found at (remove the spaces):
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1) A 180-N block rests on a frictionless inclined plane of slope angle 35°. A cord attached to the block passes over a frictionless pulley at the top of the plane and is attached to a second block. a) What must the weight of the second block be in Newtons if the system is in equilibrium? b) What is the mass of the second block in kilograms and in slugs?
1a) I got that the weight is 103 N. 2b) I got 0.07346 slug and 1.072 kg
2) A piece of angle iron is hinged as shown in (Figure 1): . Determine the resultant torque at the hinge due to the two acting forces. The top arm is 12 cm long, the bottom arm is 10 cm.
2a) I got 5.72 N x m
3) What force T at an angel of 30° above the horizontal is required to drag a 60-lb chest to the right at constant speed if μk=.22?
3a) I got tension = 17
4) Compute the center of gravity of the barbell system drawn in Figure 3. Assume the weight of the 36-in connecting rod is negligible.
4a) I got the center of gravity as being 13.8 in.
Let's go through each problem and check your answers one by one:
1) In this problem, a block is resting on an inclined plane with an angle of 35°. A cord attached to the block passes over a pulley and is connected to a second block. We need to find the weight of the second block when the system is in equilibrium.
To solve this problem, we can start by drawing a free-body diagram for each block and applying Newton's second law. The forces acting on the first block are its weight (mg) and the normal force (N). The forces acting on the second block are its weight (m2g) and tension in the cord (T).
Since the system is in equilibrium, the net force on each block must be zero. In the vertical direction, we can set up the following equation for the first block:
mg - N = 0, which implies N = mg.
For the second block, we have:
T - m2g = 0, which implies T = m2g.
We also need to consider the relationship between the masses of the two blocks. Since the cord is assumed to be massless, the tension in the cord is the same for both blocks. Therefore, T = m1g = m2g, which implies m1 = m2.
From the information given, we know the weight of the first block is 180 N. Therefore, the weight of the second block (m2g) is also 180 N.
a) So, the weight of the second block must be 180 N.
b) To find the mass of the second block, we can use the equation F = mg, where F is the weight and g is the acceleration due to gravity (9.8 m/s^2).
Since the weight of the second block is 180 N, the mass (m2) can be calculated as:
m2 = F / g = 180 N / 9.8 m/s^2 ≈ 18.37 kg
To convert the mass to slugs, we need to use the conversion factor: 1 slug ≈ 14.59 kg. Therefore:
m2 in slugs ≈ 18.37 kg / 14.59 kg/slug ≈ 1.26 slugs
1a) Your answer for the weight of the second block (103 N) is incorrect. It should be 180 N.
1b) Your answer for the mass of the second block in kilograms (1.072 kg) is incorrect. It should be approximately 18.37 kg. Your answer for the mass in slugs (0.07346 slug) is also incorrect. It should be approximately 1.26 slugs.
2) In this problem, we have an angle iron hinged at a point, and we need to find the resultant torque at the hinge due to two acting forces.
To solve this problem, we need to calculate the torque produced by each force and then find their sum. The torque produced by a force is given by the equation τ = r * F * sin(θ), where r is the perpendicular distance from the axis of rotation to the line of action of the force, F is the magnitude of the force, and θ is the angle between the force and the lever arm.
Given that the top arm of the angle iron is 12 cm long and the bottom arm is 10 cm long, we can calculate the torques produced by each force using the given lengths and magnitudes of the forces. After that, we can find the sum of the torques.
2a) To find the resultant torque at the hinge, we calculate the torque produced by each force and then add them together. Based on the provided information, we cannot determine the magnitudes of the forces or the angles between the forces and the lever arms. Therefore, we cannot verify the answer you provided (5.72 N x m).
3) In this problem, we need to determine the force required to drag a 60-lb chest to the right at a constant speed, given a coefficient of kinetic friction (μk) of 0.22 and an angle of 30° above the horizontal.
To solve this problem, we can start by drawing a free-body diagram for the chest, considering the forces acting on it. The forces acting on the chest are its weight (mg), the normal force (N), and the frictional force (f) opposing its motion.
Since the chest is moving to the right at a constant speed, we know that the force due to friction is equal in magnitude and opposite in direction to the applied force. Therefore, the force due to friction can be calculated using the equation f = μk * N.
To find the required force at an angle above the horizontal, we can use the equation F = ma, where F is the required force, m is the mass, and a is the acceleration.
Given that the weight of the chest is 60 lb, we can convert it to Newtons using the conversion factor: 1 lb ≈ 4.45 N.
3a) To find the required force (T), we start by calculating the normal force (N) from the weight of the chest:
Weight = mg, which implies N = mg.
Now we can calculate the frictional force (f) using the coefficient of kinetic friction (μk) and the normal force (N):
f = μk * N ≈ 0.22 * mg
Since the frictional force is equal in magnitude and opposite in direction to the applied force (T), we have:
f = T, which implies T ≈ 0.22 * mg.
Given that the weight of the chest is 60 lb ≈ 60 * 4.45 N, we can substitute this value into the equation:
T ≈ 0.22 * 60 * 4.45 N ≈ 17 N.
Therefore, the tension required to drag the chest to the right at a constant speed is approximately 17 N.
4) In this problem, we need to compute the center of gravity of a barbell system shown in Figure 3. The weight of the 36-in connecting rod is assumed to be negligible.
To find the center of gravity, we need to consider the weights and locations of the individual components. The center of gravity is the average position of the distributed weight.
4a) The center of gravity can be found by considering the weight distribution and calculating the weighted average position. Based on the provided information, we cannot determine the weights or positions of the components in the barbell system. Therefore, we cannot verify the answer you provided (13.8 in).
I hope this explanation helps you in verifying your answers. If you have any further questions or need more clarification, please feel free to ask!