Recent content by lion_

The picture you see is the rod being released from rest at that angle.

But at the instant theta is 30 degrees it is released from rest. So imagine a hand holding the rod and as soon as it is released the rod has no initial angular velocity. The normal component of the acceleration is obviously 0 at that instant but there is tangential acceleration of the rod.

Let me make this a little bit clearer the rod is released from rest at 30 degrees. Now I don't have to worry about horizontal motion.

Homework Statement Given: ##\mu_B=0.52## ##\theta=30^{\circ}## Weight- ##25## lb ##\omega=0## ## l=6## ft ##r_c=3\sqrt 2## radius of curvature. Homework Equations My Equations of motion are the following: ##\xleftarrow{+}\sum F_x=N_A\sin 60 - F_B=0## ##\downarrow{+} \sum F_y= N_A \cos...

I figured it out it was a little bit of common sense. First I pictured the plane flying horizontally the normal force and weight are parallel. If the plane tilts 90 degrees, then the normal force is perpendicular to the weight. So then some advanced geometry tells me if it shifts 15 degrees to...

Image didn't come out like I wanted too

Homework Statement The airplane traveling at a constant speed of 50m/s is executing a horizontal turn. If the plane is banked at ##\theta=15## when the pilot experiences only a normal force on the seat of the plane, determine the radius of curvature of the turn. Also what is the normal force of...

Homework Statement When a force of 2 lb is applied to the handles of the brad squeezer, it pulls in the smooth rod AB. Determine the force P exerted on each of the smooth brads at C and D Homework Equations ##\sum F_x=0## ##\sum M=0## ##\sum F_y=0## The Attempt at a Solution I drew a free...

Hope this helps

Try redrawing your triangle a little differently so that the angle is 135 degrees from the horizontal. Imagine flipping your triangle upside down.

So you need to calculate ##v_f=\sqrt{v_x^2+v_y^2}=\sqrt{14^2\cos^2 41^{\circ}+(18.14)^2}=20.99## I guess that what was missing thanks.

How is the horizontal component needed? The only thing we are concerned with is the height, thus once the snowball reaches maximum height, it becomes a one dimensional problem regardless of how fast the horizontal component is traveling. It is independent of gravity.

Homework Statement A ##1.50## kg snowball is fired from a cliff ##12.5##m high with an initial velocity of ##14.0## m/s, directed ##41^{\circ}## above the horizontal. How fast does the ball travel when it hits the ground. Homework Equations ##E=K+U## ##V^2_f=v^2_i+2a(y_f-y_i)## ##y_f=v_i \sin...

If the spring is compressed, isn't potential energy for the spring negative?

Homework Statement A spring constant of ##3200 N/m## is initially streched until the elastic potential energy is ##1.44## J (##U=0## for no stretch). What is the change in elastic potential energy if the initial stretch is changed to (a) a stretch of ##2.0## cm, (b) a compression of ##2.0## cm...