Answered

Looking for trustworthy answers? Westonci.ca is the ultimate Q&A platform where experts share their knowledge on various topics. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.

A mechanic pushes perpendicularly on the end of a 0.15m wrench. If the
mechanic applies a force of 230N counter clockwise, what is the torque applied
on the bolt?

Sagot :

Answer:

τ = 34.5 N.m

Explanation:

The force applied by the mechanic is 230 N.

Thus, F = 230 N

Now, this force is applied in a counterclockwise manner while pushing perpendicularly at the end of the 0.15m long wrench.

Thus, r = 0.15 m

Formula for the torque in this case is;

τ = Fr sin θ

θ = 90° since mechanic pushed perpendicularly.

Thus;

τ = 230 × 0.15 × sin 90

τ = 34.5 N.m

Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. We're glad you chose Westonci.ca. Revisit us for updated answers from our knowledgeable team.