Welcome in Collection of Solved Problems in Physics. The Collection contains tasks at various level in mechanics, electromagnetism, thermodynamics and optics. Mechanics 1 – Revision notes 1. Kinematics in one and two dimensions. Force on the pulley Resultant force Acting at 45º to the horizontal R = 17642 + 17642. Collection of Solved Problems in Physics. A bucket with mass m 2 and a block with mass m 1 are hung on a pulley system. Mechanics of rigid body (16.

• Pulley Practice Problems
• Solved Problems In Mechanics

Pulley Problems On this page I put together a collection of pulley problems to help you understand pulley systems better. The required equations and background reading to solve these problems are given on the, the, and. Problem # 1 A block of mass m is pulled, via pulley, at constant velocity along a surface inclined at angle θ. The coefficient of kinetic friction is μ k, between block and surface. Determine the pulling force F.

Answer: mgcos θ μ k+ mgsin θ Problem # 2 Two blocks of mass m and M are hanging off a single pulley, as shown. Determine the acceleration of the blocks. Ignore the mass of the pulley. Problem # 3 Two blocks of mass m and M are connected via pulley with a configuration as shown.

The coefficient of static friction is μ s, between block and surface. What is the maximum mass m so that no sliding occurs? Answer: maximum m = M μ s Problem # 4 Two blocks of mass m and M are connected via pulley with a configuration as shown.

The coefficient of static friction is μ s, between block and surface. What is the minimum and maximum mass M so that no sliding occurs? Problem # 5 Two blocks of mass m and M are connected via pulley with a configuration as shown. The coefficient of static friction is μ s, between blocks and surface. What is the maximum mass m so that no sliding occurs? Answer: Maximum m = M μ s/(sin θ−cos θ μ s) Problem # 6 Two blocks of mass m and M are connected via pulley with a configuration as shown. The coefficient of static friction between the left block and the surface is μ s1, and the coefficient of static friction between the right block and the surface is μ s2.

Formulate a mathematical inequality for the condition that no sliding occurs. There may be more than one inequality. Problem # 7 A block of mass m is pulled, via two pulleys as shown, at constant velocity along a surface inclined at angle θ.

The coefficient of kinetic friction is μ k, between block and surface. Determine the pulling force F.

Ignore the mass of the pulleys. Problem # 8 A block of mass m is lifted at constant velocity, via two pulleys as shown. Determine the pulling force F.

Ignore the mass of the pulleys. Problem # 9 A block of mass M is lifted at constant velocity, via an arrangement of pulleys as shown. Determine the pulling force F. Ignore the mass of the pulleys. The hints and answers for these pulley problems will be given next. Hints And Answers For Pulley Problems Hint and answer for Problem # 2 This is called the Atwood machine and is commonly used for demonstration in physics classes. Apply Newton's second law to the block on the left.

Pulley Practice Problems

We have Mg− T = Ma (taking the downward direction as positive). Apply Newton's second law to the block on the right. We have mg− T = - ma (the acceleration of the two blocks have opposite signs, since one moves up and the other moves down). Combine these two equations and we can find an expression for the acceleration of the blocks. Answer: a = ( M− m) g/( M+ m) Hint and answer for Problem # 4 For the maximum mass M, the block is on the verge of sliding down the incline. This means that Mgsin θ− T− Mgcos θ μ s = 0, where T is the tension in the rope.

Since T = mg, we can calculate the maximum M from the previous equation. For the minimum mass M, the block is on the verge of sliding up the incline. This means that Mgsin θ− T+ Mgcos θ μ s = 0, where T = mg.

We can calculate the minimum M from the previous equation. Answer: Minimum M = m/(sin θ+cos θ μ s), Maximum M = m/(sin θ−cos θ μ s) Hint and answer for Problem # 6 This is a challenging problem! It took me a while to figure this one out! Jquery serialize json. At some angle θ 1 θ max1 block M will slide down on its own if there is no rope attached. Similarly, at some angle θ 2 θ max2 block m will slide down on its own if there is no rope attached.

It is known that θ max1 = atan( μ s1) and θ max2 = atan( μ s2). If θ 1 ≤ θ max1 and θ 2 ≤ θ max2 then no sliding occurs. There are three more cases to consider. Case 1: θ 1 θ max1 and θ 2 ≤ θ max2.

Solved Problems In Mechanics

Apply the equilibrium equation to block M in which it is on the brink of sliding down. We have: Mgsin θ 1− Mgcos θ 1 μ s1− T min1 = 0, where T min1 corresponds to the minimum rope tension preventing block M from sliding down.

(Note that the system naturally 'settles' such that the rope tension T required to stop the block from sliding down is the minimum possible amount). For T θ max2.

This is the same as case 1, by symmetry. Hence, the final inequality for this case is: msin θ 2− mcos θ 2 μ s2 ≤ Msin θ 1+ Mcos θ 1 μ s1 Case 3: θ 1 θ max1 and θ 2 θ max2.

The blocks will slide together in one direction or the other. To determine the direction we must first calculate the net force pulling down on each block along their respective inclines, as a result of gravity. We do this as follows: For block M, F net1 = Mgsin θ 1− Mgcos θ 1 μ s1. And F net1 0 since θ 1 θ max1. For block m, F net2 = mgsin θ 2− mgcos θ 2 μ s2. And F net2 0 since θ 2 θ max2.

We now have three sub-cases to consider. The final inequalities for this case will be given within these three sub-cases, as follows. Case 3A: If F net1 = F net2 the blocks will not slide. Case 3B: If F net1 F net2, then F net1 ≤ mgsin θ 2+ mgcos θ 2 μ s2 for no sliding. Note that F net1 is equal to the rope tension, and this rope tension is the minimum required to prevent block M from sliding down the incline.

Hence, for no sliding: Msin θ 1− Mcos θ 1 μ s1 ≤ msin θ 2+ mcos θ 2 μ s2 Case 3C: If F net2 F net1, then F net2 ≤ Mgsin θ 1+ Mgcos θ 1 μ s1 for no sliding. Note that F net2 is equal to the rope tension, and this rope tension is the minimum required to prevent block m from sliding down the incline. Hence, for no sliding: msin θ 2− mcos θ 2 μ s2 ≤ Msin θ 1+ Mcos θ 1 μ s1 We are done!

Hint and answer for Problem # 7 Apply the condition of static equilibrium to the block. We have 2 F− mgsin θ− mgcos θ μ k = 0. The term 2 F comes from a force analysis in which we see that there are two segments of rope pulling equally on the block. We then solve this equation for F. Answer: F = (1/2) mg(sin θ+ μ kcos θ) Hint and answer for Problem # 8 Apply the condition of static equilibrium to the block. We have 2 F− mg = 0.

The term 2 F comes from a force analysis in which we see that there are two segments of rope pulling equally on the block. We then solve this equation for F. Answer: F = mg/2 Hint and answer for Problem # 9 Upon close inspection we see that the bottom two pulleys are held up by four segments of rope. The tension in the rope is assumed equal throughout its length (a good assumption for ropes in general since they weigh little).

Three of the four rope segments are vertical while the remaining rope segment is at a small angle with the vertical. But for ease of calculation we can treat it as being exactly vertical. Since we are ignoring the mass of the pulleys, the tension in the four rope segments must equal the weight of the mass, in order to satisfy the condition of static equilibrium. Hence, 4 F− Mg = 0. We then solve this equation for F. Answer: F = Mg/4 Bonus Problem A conveyor belt carrying aggregate is illustrated in the figure below.

A motor turns the top roller at a constant speed, and the remaining rollers are allowed to spin freely. The belt is inclined at an angle θ. To keep the belt in tension a weight of mass m is suspended from the belt, as shown. Find the point of maximum tension in the belt. You don’t have to calculate it, just find the location and give a reason for it. You can get the solution for this in PDF format. It's available through.