Tipping a crate friction physics problem

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To determine the force needed to tip a 1010 N crate by pulling at a 53-degree angle, the calculation involves balancing the torques. The counterclockwise torque generated by the pull must equal the clockwise torque due to the weight of the crate. The friction force must also be considered, as it prevents slipping, and the minimum coefficient of static friction can be calculated based on the forces involved. The discussion highlights the need for clarity in torque calculations and the importance of understanding the forces acting on the crate. Accurate calculations are essential for solving this physics problem effectively.
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A worker wants to turn over a uniform 1010 N rectangular crate by pulling at 53.0 degrees on one of its vertical sides. The floor is rough enough to prevent the crate from slipping.
YF-11-49.jpg


1. What pull is needed to just start the crate to tip?
2. How hard does the floor push on the crate?
3. Find the friction force on the crate.
4. What is the minimum coefficient of static friction needed to prevent the crate from slipping on the floor?

1. torque = (F_pull)cos(270-53)(1.5m) = 1010 * 1.5 = 1515 N
the problem is getting started. i don't think i did this right.
 
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. torque = (F_pull)cos(270-53)(1.5m) = 1010 * 1.5 = 1515 N
This is the counterclockwise torque. What is the clockwise torque?
 
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