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PRACTICE IT Use the worked example above to help you solve this problem. (a) A man applies a force of F= 3.00 x 1o? N at an angle of 60.0" to a door, x = 2.50 m from the hinges. Find the torque on the door, choosing the position of the hinges as the axis of rotation. N-m (b) Suppose a wedge is placed 1.50 m from the hinges on the other side of the door. What minimum force must the wedge exert so that the force applied in part (a) won't open the door?

PRACTICE IT Use the worked example above to help you solve this problem. (a) A man applies a force of F= 3.00 x 1o? N at an angle of 60.0" to a door, x = 2.50 m from the hinges. Find the torque on the door, choosing the position of the hinges as the axis of rotation. N-m (b) Suppose a wedge is placed 1.50 m from the hinges on the other side of the door. What minimum force must the wedge exert so that the force applied in part (a) won't open the door?

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Please answer both practice and exercise parts.
GOAL Apply the more general definition of torque. 300 N Hinge 60.0° 2,00 m Hinge 260 N 150 N 2.00 m (a) Top view of a door being pushed by a 300- N force. (b) The components of the 300-N force. PROBLEM (a) A man applies a force of F= 3.00 x 102 N at an angle of 60.0° to the door of Figure (a), 2.00 m from well-oiled hinges. Find the torque on the door, choosing the position of the hinges as the axis of rotation. (b) Suppose a wedge is placed 1.50 m from the hinges on the other side of the door. What minimum force must the wedge exert so that the force applied in part (a) won't open the door? STRATEGY Part (a) can be solved by substitution into the general torque equation. In part (b) the hinges, the wedge, and the applied force all exert torques on the door. The door doesn't open, so the sum of these torques must be zero, a condition that can be used to find the wedge force. SOLUTION (A) Compute the torque due to the applied force exerted at 60.0". Substitute into the general torque TE = rFsin e = (2.00 m)(3.00 x 102 N) sin 60.0° - (2.00 m)(2.60 x 102 N) = 5.20 x 102 N.m equation. (B) Calculate the force exerted by the wedge on the other side of the door. Set the sum of the torques equal Thinge + Twedge+ TE =0 to zero. The hinge force provides no 0+ Fwedge(1.50 m) sin (-90.0) + 5.20 x 102 N m torque because it acts at the axis (r = 0). The wedge force acts at an angle of -90.0, opposite the Fwedge" 347 N %3D upward 260 N component. LEARN MORE REMARKS Notice that the angle from the position vector to the wedge force is -90°. This is because, starting at the position vector, it's necessary to go 90° clockwise (the negative angular direction) to get to the force vector. Measuring the angle in this way automatically supplies the correct sign for the torque term and is consistent with the right-hand rule. Alternately, the magnitude of the torque can be found and the correct sign chosen based on physical intuition. Figure (b) illustrates the fact that the component of the force perpendicular to the lever arm causes the torque. QUESTION To make the wedge more effective in keeping the door closed, should it be placed closer to the hinge or to the doorknob? O closer to the hinge O closer to the doorknob PRACTICE IT Use the worked example above to help you solve this problem. (a) A man applies a force of F = 3.00 x 102 N at an angle of 60.0° to a door, X = 2.50 m from the hinges. Find the torque on the door, choosing the position of the hinges as the axis of rotation. N-m (b) Suppose a wedge is placed 1.50 m from the hinges on the other side of the door. What minimum force must the wedge exert so that the force applied in part (a) won't open the door? EXERCISE HINTS: GETTING STARTED | Il'M STUCK! A man ties one end of a strong rope 8.60 m long to the bumper of his truck, 0.587 m from the ground, and the other end to a vertical tree trunk at a height of 2.90 m. He uses the truck to create a tension of 8.08 x 102 N in the rope. Compute the magnitude of the torque on the tree due to the tension in the rope, with the base of the tree acting as the reference point. N-m
Transcribed Image Text:GOAL Apply the more general definition of torque. 300 N Hinge 60.0° 2,00 m Hinge 260 N 150 N 2.00 m (a) Top view of a door being pushed by a 300- N force. (b) The components of the 300-N force. PROBLEM (a) A man applies a force of F= 3.00 x 102 N at an angle of 60.0° to the door of Figure (a), 2.00 m from well-oiled hinges. Find the torque on the door, choosing the position of the hinges as the axis of rotation. (b) Suppose a wedge is placed 1.50 m from the hinges on the other side of the door. What minimum force must the wedge exert so that the force applied in part (a) won't open the door? STRATEGY Part (a) can be solved by substitution into the general torque equation. In part (b) the hinges, the wedge, and the applied force all exert torques on the door. The door doesn't open, so the sum of these torques must be zero, a condition that can be used to find the wedge force. SOLUTION (A) Compute the torque due to the applied force exerted at 60.0". Substitute into the general torque TE = rFsin e = (2.00 m)(3.00 x 102 N) sin 60.0° - (2.00 m)(2.60 x 102 N) = 5.20 x 102 N.m equation. (B) Calculate the force exerted by the wedge on the other side of the door. Set the sum of the torques equal Thinge + Twedge+ TE =0 to zero. The hinge force provides no 0+ Fwedge(1.50 m) sin (-90.0) + 5.20 x 102 N m torque because it acts at the axis (r = 0). The wedge force acts at an angle of -90.0, opposite the Fwedge" 347 N %3D upward 260 N component. LEARN MORE REMARKS Notice that the angle from the position vector to the wedge force is -90°. This is because, starting at the position vector, it's necessary to go 90° clockwise (the negative angular direction) to get to the force vector. Measuring the angle in this way automatically supplies the correct sign for the torque term and is consistent with the right-hand rule. Alternately, the magnitude of the torque can be found and the correct sign chosen based on physical intuition. Figure (b) illustrates the fact that the component of the force perpendicular to the lever arm causes the torque. QUESTION To make the wedge more effective in keeping the door closed, should it be placed closer to the hinge or to the doorknob? O closer to the hinge O closer to the doorknob PRACTICE IT Use the worked example above to help you solve this problem. (a) A man applies a force of F = 3.00 x 102 N at an angle of 60.0° to a door, X = 2.50 m from the hinges. Find the torque on the door, choosing the position of the hinges as the axis of rotation. N-m (b) Suppose a wedge is placed 1.50 m from the hinges on the other side of the door. What minimum force must the wedge exert so that the force applied in part (a) won't open the door? EXERCISE HINTS: GETTING STARTED | Il'M STUCK! A man ties one end of a strong rope 8.60 m long to the bumper of his truck, 0.587 m from the ground, and the other end to a vertical tree trunk at a height of 2.90 m. He uses the truck to create a tension of 8.08 x 102 N in the rope. Compute the magnitude of the torque on the tree due to the tension in the rope, with the base of the tree acting as the reference point. N-m
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