00:01 So in this problem, we're given two masses attached together by a weightless string on a slope.
00:07 And we're asked to calculate some properties of the system, given our understanding of kinematics and static forces.
00:15 So the first part of the question is telling us that given the system is initially at rest, write down newton's first law and use it to calculate the magnitude of the tension in the string.
00:26 So newton's first law is to tell us that if an object is in motion or not in motion, it will either continue in motion or remain stationary unless acted upon by an external force.
00:40 So assuming that the system is remaining stationary, what we're saying is the forces acting on the mass one must be equal.
00:48 So what do we have acting on mass one? well, mass one obviously has its weight going down and we have the tension in the string pulling upwards.
00:59 And so what newton's first law is telling us is that the weight has to be equal to the tension in the string.
01:07 Ring and the weight of the thing of the mass is easily calculated as the mass times g which is about 15 .7 newtons in this case and so that's our 11 part answer to part a now for part b we're told that the static friction force is at its maximum possible value and we're asked to to draw a free body diagram showing the forces acting on the second mass and use this to calculate the second mass.
01:48 So let's draw a slope.
01:58 And then we're told this angle here, theta is equal to 35 degrees.
02:07 But i'm just gonna leave it labeled as theta.
02:09 And so we have our mass, m2, and the string pulls up to the pulley and then goes over the edge, but we just need to focus on the forces here.
02:28 So we have the tension in the string this way, and now we have the weight, the mass of the object, which acts from its centre of mass, down, which is m2g, and then we also have a normal force that acts perpendicular to the surface, this.
02:59 And finally, we also have the force of static friction, which acts in a direction to oppose the motion of the block.
03:09 And so these are the four forces acting on the system.
03:12 And because we're assuming that the system is in equilibrium, we need to balance these forces and use that condition to determine what the mass m2 must be.
03:22 So the way we're going to do this is we're going to think about the mass, the weight of the mass, mg, being resolved, m2g, being resolved into two components, one perpendicular and one parallel to the slope.
03:47 If we draw this line and this line, we can show through analysing our angles that this angle here is also theta, assuming that this direction here is parallel to the slope, and this is perpendicular.
04:09 And so what we say is that the weight that's parallel to the slope is equal to the weight that is parallel to the slope is equal to m to g.
04:35 This leg of the triangle is given by the sine, the opposite over the hypotenuse, sine theta, and then the weight perpendicular is given by m2g cosine of theta.
04:55 Now, the action of the weight perpendicular to the slope must be balanced by the normal force.
05:03 So we know that the normal force is actually just equal to m2g cosine of theta.
05:16 And now the force is acting parallel to the slope, we have the tension, which must be equal to the weight parallel to the slope, m2 g sine theta, plus the static friction force.
05:35 We also know that the static friction force can be written as the coefficient of static static friction multiplied by the normal force.
05:44 And so we can write the tension is equal to the mass of the second block times g times sine of theta plus the coefficient of static friction times the normal force, which we've already said is the mass of the second object times g cosine of theta...
