3. A 50 cm wrench is rotated with a torque of magnitude 115...

3. A 50 cm wrench is rotated with a torque of magnitude 115 N·m. A force is applied 25 cm from the fulcrum, at angle of 20°. a. What is the magnitude of the force? b. Is 115 N·m the largest torque that can be generated using a force of this magnitude and this wrench? How could the force be applied to generate the maximum torque? c. If the same force is still applied 25 cm from the fulcrum what angle will produce double the original torque?

Transcript

00:03 Hello, here the range has a total length of 50 centimeters and it's rotated with a torque that has a magnitude of 15 newton meter and the force is applied at a distance of 25 centimeters from the fore column at an angle of 20 degrees.

00:30 Now we have to in question one we have to find magnitude of this force f1 so let's illustrate it so that is a forculum and this torque this range has an length of l but the force is applied at distance d1 and this force f1 makes an angle alpha therefore this torque t1 is f1 d1 sign alpha therefore f1 is torque 1 over d1 sign alpha that is 175 newtons now let's answer question 2 in question 2 we have to determine whether this is the largest force whether this is the largest torque that can be generated using the force of the same magnetion.

02:28 So now we have to calculate basically maximum torque, which can be created.

02:34 And the maximum torque will be created when the force of f1 is applied at distance l from the 4 column.

02:43 So therefore, the maximum torque equals to f1 times l.

02:48 And also that has to be applied perpendicular that's why that's multiplied by sign of 90 degree that is 175 newtons times half a meter times one that is 87 and a half newton meter which is roughly 88 newton meter so that is answer to question now let's answer question 3 so basically we've already answered it so it means that here f1 has to be applied at el from the fculum perpendicular to the range okay so now the same force is applied now we have to calculate the angle at which the same force applied at distance of 25 centimeters will produce the double torque so let's label this torque as torque t3 so that equals to 2 torque 1 and we have to calculate angle here 2 times torque 1 equals to f1 times d1 multiplied by sign of angle alpha 3 therefore sign of alpha 3 equals to 2 times 15 newton meter divided by force f1 which is 175 newtons and by distance d1 which is 25 centimeters and by distance d1 which is 25 centimeters let's calculate the sign...

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