9.) A student is attempting to loosening a bolt on their car but is struggling.
ID: 3111873 • Letter: 9
Question
9.) A student is attempting to loosening a bolt on their car but is struggling. Eventually, they apply their entire 700 N of weight on the lug wrench which is a quarter of a meter long. Due to the current position of the bolt, the student can only press down on the wrench which is angled 30 degrees above horizontal. What is the magnitude of the torque applied to the bolt?9. A student is attempting to loosening a bolt on their car but is struggling. Eventually, they apply their entire 700 N of weight on the lug wrench which is a quarter of a meter long. Due which is to the current position of the bolt, the student can only press down on the wrench angled 30° degrees above horizontal. What is the magnitude of the torque applied to the bolt? 10. A student is about to open a jar and are going to apply force to the top of the jar with a vector of F = 200i +200j. Since the jar is a cylinder, the choice of position vector is up to the student. What is the vector with the maximum torque that the student can apply to the top of the jar? (Hint: consider the position vector based on the formula for the magnitude of the torque and what angle it should have with the force vector)
Explanation / Answer
Q9) Torque = |rXF|
angle made by student = 30 degrees
Length of the wrench = 1/4m = 0.25m
Force applied = 700N
Torque = |F||r| sin(theta) = |700||1/4| sin(30) = 700 * 1/4 * 1/2 = 87.5 Nm
10) Force is inclined to an angle of 45 degrees from the horizontal since tan(theta) = 1
Now the student should use the torque formula
torque = |Fxr| = |F||r| sin(theta)
r doesn't matter since the distance remains and force remains same, the only thing matter is the angle, so for maximizing sin(theta) we should keept it at 90 degrees
so the direction will be 135 to the horizontal to make it 90 degrees with the force
Torque = 200sqrt 2 * (cylinder dimensionn)
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