Introductory Physics 1 - Test 3 (Torque & Equilibrium)

Air density = 1.23 kg/m³ 1/2 × 1.23 × 0.5 m × 0.9 m × (17 m/s)² = 79.98 τ = 79.98 × 1.2 m = 95.98 N-m

An 80 kg man stands in a very strong wind moving at 17 m/s at torso height. As you know, he will need to lean in to the wind, and we can model the situation to see why. Assume that the man has a mass of 80 kg, with a center of gravity 1.0 m above the ground. The action of the wind on his torso, which we approximate as a cylinder 50 cm wide and 90 cm long centered 1.2 m above the ground, produces a force that tries to tip him over backward. To keep from falling over, he must lean forward. a) What is the magnitude of the torque provided by the wind force? Take the pivot point at his feet. Assume that he is standing vertically. Assume that the air is at standard temperature and pressure.

a) 60/0.04 × 0.02 = 30 N b) When the ball is released at the extreme point, the ball is moving towards the mean position due to the restoring force created in the ligament. Speed is also increasing towards the mean position whenever this happens, so the net force is acting upward towards the mean position.

a) When the ligament is stretched, it exerts a restoring force. A stretch of 4 cm produces a restoring force of 60 N. What is the restoring force for a 2 cm stretch? b) When the ball hangs from the ligament, it stretches a certain amount. If the ball is then pulled downward, the ligament experiences a greater stretch. When the ball is subsequently released, what is the direction of the net force on the ball?

T = F x r x sin(q) 1. θ = 90, Sin(θ) = 1, τ(1) = FL 2. θ = 90, Sin(θ) = 1, τ(2) = F(L/2) 3. 90 < θ < 180, 1 < sin(θ) < -1 but ≠ 1, so τ(3) < τ(4) 4. 0 < θ < 90, 0 < sin(θ) < 1 but ≠ 1, so τ(4) < τ(1) 5. θ = 0, sin(θ) = 0; τ(5) = 0

Torque: As shown in the figure, a given force is applied to a rod in several different ways. In which case is the torque about the pivot P due to this force the greatest?

The first force (applied perpendicular to the door) has a torque of F × r while the second torque force is F(sin 60) × r

Torque: Two equal-magnitude forces are applied to a door at the doorknob. The first force is applied perpendicular to the door, and the second force is applied at 30° to the plane of the door. Which force exerts the greater torque about the door hinge?

a) (3.2)(9.8)(50/2) / 5 F(tendon) = 156.8 N b) 156.8 - (3.2)(9.8) F(knee) = 125.44 N c) Vertically downward

If you stand on one foot while holding your other leg up behind you, your muscles apply a force to hold your leg in this raised position. We can model this situation as in (Figure 1). The leg pivots at the knee joint, and the force that holds the leg up is provided by a tendon attached to the lower leg as shown. Assume that the lower leg and the foot have a combined mass of 3.2kg, and that their combined center of gravity is at the center of the lower leg. a) How much force must the tendon exert to keep the leg in this position? b) As you hold your leg in this position, the upper leg exerts a force on the lower leg at the knee joint. What is the magnitude of this force? c) What is the direction of the force exerted by the upper leg on the lower leg at the knee joint?

F(chain) × 9.1 cm = F(paddle) × 17 cm F(paddle) = (9800 N × 9.1 cm)/ 17 cm F(paddle) = 5245.88 N

A bike chain can support a tension of no more than 9800 N. The pedal connects to a crank 17 cm from the axle, and the gear pulling the chain has a 9.1 cm radius. When riding at a constant speed, with the crank and pedal horizontal, as in (Figure 1), what is the maximum force that can be applied to the pedal before the chain breaks?

N(front) = 0.60 W N(rear) = 0.40 W 0.60(1.8 + 0.5 - d) - 0.4(d - 0.5) = 0 = 1.08 + 0.3 - 0.60d - 0.4d + 0.2 = 0 d = 1.58 m

A hippo's body is 4.0 m long with front and rear feet located as in (Figure 1). The hippo carries 60% of its weight on its front feet. How far from its tail is the hippo's center of gravity?

T = 8.6 kg × 9.8 m/s² = 84.28 kg*m/s² = 84.28 × 2 = 168.56 N

A vendor hangs an 8.6 kg sign in front of his shop with a cable held away from the building by a lightweight pole. The pole is free to pivot about the end where it touches the wall, as shown in (Figure 1). What is the tension in the cable?

a) 130 × 9.8 × 0.25/ 1.1 = 289.55 N b) 289.55/ 9.8 × 130 = .227

A woman is pushing a load in a wheelbarrow, as in (Figure 1). The combined mass of the wheelbarrow and the load is 130kg, with a center of gravity at d = 0.25m behind the axle. The woman supports the wheelbarrow at the handles, 1.1 m behind the axle. a) What is the force required to support the wheelbarrow? b) What fraction of the weight of the wheelbarrow and the load does this force represent?

a) (570)(54)/ 76 = 405 405 / 2 = 202.5 N b) 570 - 405 = 165 165 / 2 = 82.5 N

A woman weighing 570 N does a pushup from her knees, as shown in the figure. (Figure 1) a) What are the normal forces of the floor on each of her hands? b) What are the normal forces of the floor on each of her knees?

a) T(L) + T(R) = 500 T(L) + 130 = 500 T(L) = 500 - 130 T(L) = 370 N b) -(400 N)(2 m) - (100 N)(5) + T(R)(10) = 0 -800 - 500 + T(R)(10) = 0 T(R)(10) = 1300 T(R) = 130 N

Equilibrium: A 10-m uniform beam weighing 100 N is supported horizontally by two vertical ropes at its ends. a) If a 400-N person sits at a point 2.0 m from the left end of the beam, what is the tension in the left rope? b) If a 400-N person sits at a point 2.0 m from the left end of the beam, what is the tension in the right rope?

(M x (61.3 - 0) / (100))((100 - 61.3 / 2) + (50)(90 - 61.3) Solving for "M", we obtain 127 g

Equilibrium: A meter stick balances at the 50.0-cm mark. If a mass of 50.0 g is placed at the 90.0-cm mark, the stick balances at the 61.3-cm mark. What is the mass of the meter stick?

500 × 1.5 = 350 × (x) x = 500 × 1.5 / 350 = 2.14 m

Equilibrium: A uniform 40-N board supports two children weighing 500 N and 350 N. If the support is at the center of the board and the 500-N child is 1.5 m from its center, how far is the 350-N child from the center?

a) F(A) = 2511.25 - 82 kg × 9.8 m/s F(A) = 1707.65 N or 1.7 kN in the downwards direction b) F(B) × 1.6 m = 82 kg × 9.8 m/s × 5 m F(B) = 2511.25 N or 2.5 kN in the upwards direction

Equilibrium: An 82-kg diver stands at the edge of a light 5.0-m diving board, which is supported by two vertical pillars that are 1.6 m apart, as shown in the figure. a) Find the force exerted by pillar A. b) Find the force exerted by pillar B.

FL = FL (5800)x = (6500)(3.2 - x) 5800x = 20800 - 6500x x = 1.69 m

Equilibrium: To determine the location of the center of mass (or center of gravity) of a car, the car is driven over a scale on a horizontal floor. When the front wheels are over the scale, the weight recorded by the scale is 5800 N, and when the rear wheels are over the scale, the scale reads 6500 N. The distance between the front and rear wheels is measured to be 3.20 m. How far behind the front wheels is the center of mass located?

F = kx k = -F/x = -1.5 × 10⁻⁹ N / 5 × 10⁻⁹ m = 0.3 N/m

Experiments using "optical tweezers" measure the elasticity of individual DNA molecules. For small enough changes in length, the elasticity has the same form as that of a spring. A DNA molecule is anchored at one end, then a force of 1.5 nN (1.5 × 10⁻⁹ N) pulls on the other end, causing the molecule to stretch by 5.0 nm (5.0 × 10⁻⁹ m). What is the spring constant of that DNA molecule?

τ = 50 × 0.6 = 30 N-m

To exercise, a man attaches a 6.0 kg weight to the heel of his foot. When his leg is stretched out before him, what is the torque exerted by the weight about his knee, 50 cm away from the weight? Use g = 10 m/s².

The torque applied to the wrench is the force exerted at the end with a perpendicular component at an unknown angle that when multiplied by 0.32 m will provide 15 Nm of torque. In this case, the perpendicular component is the sine of angle. Therefore, 15 Nm = (95 N) (sin(Θ)) (0.32 m). Solving for Θ we get, sin(Θ) = 15 Nm / (95 N) (0.32 m) [Notice that the units cancel out as they should]. sin(Θ) = 15 / 30.4 = 0.4934. Θ = arcsin(0.4934) = 29.56 = 30°

Torque: A 95-N force exerted at the end of a 0.32-m long torque wrench produces a torque of 15 N ∙ m. What is the angle (less than 90°) between the wrench handle and the direction of the applied force?

τ = r × F F = (8.0)(9.8) = 78.4 N torque = (0.55)(78.4) = 43.12 N-m

Torque: A man in a gym is holding an 8.0-kg weight at arm's length, a distance of 0.55 m from his shoulder joint. What is the torque about his shoulder joint due to the weight if his arm is horizontal?

The right hand rule for torque states that if the figures of the right hand are curled from the direction of r toward the direction of F, then the thumb points in the direction of the torque. The radius vector will be along the radial direction and force C. On taking the cross product of these using the right hand thumb rule, it will make the rod move in counter-clockwise direction.

Torque: Five forces act on a rod that is free to pivot at point P, as shown in the figure. Which of these forces is producing a counter-clockwise torque about point P? (There could be more than one correct choice.)

Torque is magnitude of the perpendicular force times the distance. Here in both cases the torque is same hence the closer force will have higher magnitude. So, the first force (at the midpoint) will have the greater magnitude.

Torque: Two forces produce equal torques on a door about the door hinge. The first force is applied at the midpoint of the door; the second force is applied at the doorknob. Both forces are applied perpendicular to the door. Which force has a greater magnitude?