A 15000 n crane pivots around a friction free axle at its...

A 15,000-N crane pivots around a friction-free axle at its base and is supported by a cable making a 25$^\circ$ angle with the crane ($\textbf{Fig. E11.18)}$. The crane is 16 m long and is not uniform, its center of gravity being 7.0 m from the axle as measured along the crane. The cable is attached 3.0 m from the upper end of the crane. When the crane is raised to 55$^\circ$ above the horizontal holding an 11,000-N pallet of bricks by a 2.2-m, very light cord, find (a) the tension in the cable and (b) the horizontal and vertical components of the force that the axle exerts on the crane. Start with a free-body diagram of the crane.

Transcript

00:01 Hi.

00:03 So for this problem, we have a crane here connected to the ground to this axle.

00:11 So to start, we need to know what forces we have.

00:16 So we need to draw a free by that.

00:18 Okay, let's see what forces we have.

00:21 We have the gravitational force from the block here.

00:25 We have a gravitational force of the crane itself.

00:30 We have the tension here.

00:32 And also we have force from the axel.

00:36 We do not know to which direction is this force, but we can divide this force into two directions.

00:48 We have its x component to this direction and the y component to this direction.

01:01 So in this problem, the whole system is studying, which means, the net force and the network should be equal to zero.

01:12 So that's where we're going to get started.

01:16 So first, we're going to need the net force equal to zero, which means the net force in the x direction and the net force in the y direction, they're both zero.

01:28 So let's see.

01:29 The net force in the x direction, we have fx in the positive direction.

01:35 We have the x component from the tension in the negative direction, and since this angle is 55, this angle is 25, so this angle should be 30.

01:48 This angle should be 30, because 30 plus 25, it could be 55.

01:55 So the x component of tension is going to be t cosine 30.

02:04 So positive value minus negative value should be 0.

02:07 That's our first equation.

02:09 Then in the y direction, we have small mg from the block, we have capital mg from the crane itself, and we have fy in the positive direction.

02:23 Also, our tension has its x component, which is t sine 30 degree, and it's also in the next direction.

02:34 So the forces in the y direction are going to be fy minus positive minus negative, m g minus small mg minus t sine 30 and it's equal to zero so equation one and two give us the net force equal to zero then also the system is not rotating which means the net torque is zero so let me remind you the magnitude of torque is equal to the distance from the rotating axis to the force for example for m g this is the distance, times the magnitude of the force times sine theta, where this theta is the angle between the distance and the force, for example.

03:27 For example, for our mg here, this is our theta.

03:34 However, since sine theta is equal to this angle, theta prime, sine theta and same sine theta prime, is it equal to each other.

03:48 Because the sum of them is 180.

03:52 So here, we're going to directly use this angle as our speed.

03:57 We're going to use this angle as r2.

04:02 So let's see what works we have.

04:06 So we choose this point as our rotating axis, because we do not know the value of fx and fy yet.

04:14 So by choosing this point as our rotating axis, the torque from fx and fy can be zero, it can be zero.

04:22 We do not need to care about it...