Physics: Torque, Equilibrium, and Bridge Loads
Analyzing forces, moments, and support reactions in bridge structures
Physics: Torque, Equilibrium, and Bridge Loads
Analyzing forces, moments, and support reactions in bridge structures
Physics - Grade 9-12
- 1
A 25 N force is applied perpendicular to a wrench 0.30 m from the bolt. Calculate the torque about the bolt.
Use τ = Fd when the force is perpendicular to the lever arm.
The torque is 7.5 N·m because torque equals force times perpendicular distance, so τ = 25 N × 0.30 m = 7.5 N·m. - 2
A student pushes downward with 60 N on the end of a 0.80 m meter stick pivoted at the other end. The force makes a 90° angle with the stick. What is the torque about the pivot?
The torque is 48 N·m because τ = Fd sin θ = 60 N × 0.80 m × sin 90° = 48 N·m. - 3
A 40 N force is applied to a door 0.75 m from the hinge, but the force is applied at an angle of 30° to the door surface. Calculate the torque about the hinge.
Use the sine of the angle between the force and the door.
The torque is 15 N·m because only the perpendicular component produces torque, so τ = Fd sin θ = 40 N × 0.75 m × sin 30° = 15 N·m. - 4
A uniform 6.0 m beam weighs 300 N and is supported at both ends. A 600 N person stands at the center of the beam. What upward force does each support provide?
When all loads are centered symmetrically, the support forces are equal.
Each support provides 450 N upward. The total downward force is 300 N + 600 N = 900 N, and symmetry means the two supports share the load equally. - 5
A 10.0 m bridge beam is supported at both ends. A 1200 N car is parked 4.0 m from the left support. Ignore the weight of the beam. Find the upward force from the right support.
Choose the left support as the pivot so the left support force creates no torque.
The right support force is 480 N. Taking torques about the left support gives FR × 10.0 m = 1200 N × 4.0 m, so FR = 480 N. - 6
Using the bridge in Problem 5, find the upward force from the left support.
The left support force is 720 N. The vertical forces must balance, so FL + 480 N = 1200 N, which gives FL = 720 N. - 7
A 12.0 m uniform bridge beam weighs 2400 N and is supported at both ends. A 900 N load is placed 3.0 m from the left support. Find the upward force from the right support.
The weight of a uniform beam acts at its center.
The right support force is 1425 N. Taking torques about the left support gives FR × 12.0 m = 2400 N × 6.0 m + 900 N × 3.0 m, so FR = 17100 N·m ÷ 12.0 m = 1425 N. - 8
Using the bridge in Problem 7, find the upward force from the left support.
The left support force is 1875 N. The total downward force is 2400 N + 900 N = 3300 N, so FL + 1425 N = 3300 N and FL = 1875 N. - 9
A seesaw is balanced on a pivot. A 500 N student sits 1.5 m to the left of the pivot. A second student sits 2.0 m to the right of the pivot. What must the second student's weight be for equilibrium?
Set clockwise torque equal to counterclockwise torque.
The second student's weight must be 375 N. For rotational equilibrium, 500 N × 1.5 m = W × 2.0 m, so W = 375 N. - 10
A bridge sign hangs from a horizontal support arm. The 200 N sign is attached 1.2 m from the wall. A cable pulls upward at the end of the arm, 1.5 m from the wall. If the cable force is vertical, what force must the cable provide to keep the arm in rotational equilibrium? Ignore the weight of the arm.
The cable must provide 160 N upward. Taking torques about the wall gives Fcable × 1.5 m = 200 N × 1.2 m, so Fcable = 160 N. - 11
A beam is in static equilibrium. What two conditions must be true for the beam to remain at rest without translating or rotating?
Static equilibrium requires balance in both motion types.
The net force on the beam must be zero, and the net torque on the beam must be zero. These conditions prevent both linear acceleration and angular acceleration. - 12
A 100 N force acts 0.50 m from a pivot. A 40 N force acts on the opposite side of the pivot. How far from the pivot must the 40 N force act to balance the torques?
The 40 N force must act 1.25 m from the pivot. The balancing condition is 100 N × 0.50 m = 40 N × d, so d = 1.25 m. - 13
A 16.0 m bridge is supported at both ends. A 2000 N truck is 6.0 m from the left support, and a 1000 N load is 12.0 m from the left support. Ignore the bridge weight. Find the upward force from the right support.
Add the torques from both downward loads before dividing by the bridge length.
The right support force is 1500 N. Taking torques about the left support gives FR × 16.0 m = 2000 N × 6.0 m + 1000 N × 12.0 m = 24000 N·m, so FR = 1500 N. - 14
Using the bridge in Problem 13, find the upward force from the left support.
The left support force is 1500 N. The total downward force is 2000 N + 1000 N = 3000 N, so FL + 1500 N = 3000 N and FL = 1500 N. - 15
A bridge inspector notices that the right support force increases when a heavy truck moves closer to the right end of a bridge. Explain why this happens using torque and equilibrium.
Think about how torque changes when the same force acts farther from the pivot.
The right support force increases because the truck creates more torque about the left support as its distance from the left support increases. To maintain rotational equilibrium, the right support must provide a larger upward torque, so its upward force increases.