i just checked the answer and it is
mg/3 ( √(9+cot2(φ))
we have never encounter a problem that its answer has cotangent in it. its way out of our league. can someone please just show me how to derive the answer?
i still can't figure it out
i check the answer and is this:
mg/3 [√(9+ cot2(φ))]
we have never encounter a problem that the answer is cotangent before its way out of my league can someone please just tell me how to do it?
he didnt give us the weight and is the missing force the normal force at the contact point between the table and the rod?
<a href="https://imgur.com/U6SG6En"><img src="https://i.imgur.com/U6SG6En.jpg" title="source: imgur.com" /></a>
Homework Statement
https://imgur.com/v13K6sE
a uniform rod of mass m is placed as shown, with one of its end resting on a smooth wall while 1/4 of the rod's length is sticking out of a rough table. Find the net force the table exert on the rod at the corner.
Homework Equations
i drew the free...
I wasnt in class that day and homework are usually much harder than lectures.
He gave is us this formula in class
Centripetal acc. = R * √ (ω4 + α2)
But I don't know how to use it in this problem
a disc or radius r = 16cm starts spinning from rest with a uniform angular acceleration of 8.0 rad/s^2. at what time is its tangential acceleration twice the centripetal acceleration.
i figured out the tangential acceleration is:
Atan = α/R = 8 / .16 = 50 m/s^2
and the centripetal...