Forces and Equilibrium Homework Help

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2. Typical right angle vector addition; be sure to find the direction as well as the magnitude.
4. Since there is no right angle here, we must use the law of cosines and law of sines. Refer to example problem on page 74 for the method. Draw the forces, the resultant, and the given angle, compare to Figure 4-6 on page 73. (31 N @ 19 deg from 15 N force)
5. Sketch situation in problem. Find the weight of the bucket--each end of rope must supply half this force in its upward component. (91.8 N)
6. Graphic solution means draw the vectors carefully and solve by measuring the resultant. (you can easily check by using trig).

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1. Basic vector resolution problem. Sketch the situation to make sure you use the correct trig function for each component.
3. Is the component force that makes the car want to roll down the hill greater than the braking force?
5. Two different forces on opposite ends of a stretcher. Find each upward component and combine for the total. A sketch will prove helpful.
7. Find upward component of maximum tension force and double since there are two wires. If angle between wires is 80.0, deg. angle to vertical must be half this.

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1. Phrase "constant speed" is key here.This is equilibrium, so there is no net force. Since surface is horizontal, normal force equals weight of trunk. If there is no acceleration, friction force must equal applied force. (0.227)
3. Here we do have acceleration, therefore, the applied force is more than the friction. Find how much extra force is needed to accelerate a 20.0 kg mass at the given rate. The remainder of the applied force must have been used to overcome friction. (0.230)
5. Friction force must at least equal force necessary to create acceleration given: a 2nd law problem.
7. We assume constant speed, so the friction force equals the horizontal component of the applied force. The normal force will be greater than the weight since the push is directed downward at 50 deg. Find that vertical (downward) component of the push that contributes to the normal force.
9. There are several forces at work here. Drawing a free body diagram will help. Remember, friction will be in the opposite direction of motion. Since the crate is accelerating, there is more applied force up the plane than the combined forces (friction and parallel component of weight) down the plane. (840 N)
11. Solving this problem will prove an important relationship is equilibrium situations. Since motion is at constant speed, there is no net force.

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1. Force x distance for big kid must equal force x distance for little kid. Balance the torques.
3. Where can we consider the weight of a uniform object to be located? Find the torque arms for the 50-N weight and the weight of the meterstick, and set up the balanced torque equation. Solve for the meterstick weight.
4. Draw a sketch of the situation. Write the force equation: ups equal downs. Place the pivot point at the location of one of the unknown forces (either end) to eliminate it from the torque equation. Write the torque equation measuring each distance from the end you chose as your pivot point. Solve for the unknown upward force. Substitute this value into the force equation and solve for the remaining upward force. (712 N and 580 N)
5. Draw a sketch. Write the force equation: ups equal downs. Place your pivot point at one of the unknown upward forces from the pillars. Write the torque equation measuring each distance from your pivot point. Be careful here, the pillars are not at the very end of the bridge. Solve for the unknown force. Use the force equation to find the other force. (2.07 x 10⁵ N and 2.13 x 10⁵ N)
7. Draw a sketch. Once again, a uniform object, so its weight vector is placed where? (a) Write the force equation to find the tension. Be careful, we have weights and masses mixed here. (b) Choose a convenient pivot point and write the torque equation with the torque arm of the rope tension as the unknown. (13.77N upward at 50.3 cm mark)
8. Like the bridge problem in # 5 but the supports are at the end. Draw a sketch being careful with the locations. Assume the scaffold is uniform. Measure all torque arms from your chosen pivot point. (500 N and 640 N)