Torque question?
A 9.00 m uniform beam is hinged to a vertical wall and held horizontally by a 5.00 m cable attached to the wall 4.00 m above the hinge, as shown in the figure below (Figure 1) . The metal of this cable has a test strength of 1.10 kN , which means that it will break if the tension in it exceeds that amount. What is the heaviest beam that the cable can support with the given configuration? The answer is 59kg. I got it wrong on all attempts. Can someone show me what I'm meant to do?
Physics torque problem?
for the wheels not to spin, the torque exerted by the engine cannot exceed the torque generated by friction; if each wheel supports 1/4 the weight of the car, the torque due to friction on each wheel is torque = force of friction x lever arm force of friction = us mg/4 = 0.8*1400kg*9.8m/s/s / 4 = 2744N the lever arm is the distance from the ground to the center of the wheel, or 0.305m, so the torque due to friction is 2744N*0.305m=836.9Nm and the engine cannot generate more torque than this
How do I relate motorcycle torque with the pulling force of an object? I want to prove that a motorcycle can tow a 100 kg object on a 0 to 25 degree slope.
The other answerers apparently have not taken freshman physics.To move an object, the force moving it has to exceed the force opposing it.The force moving it is the torque times tire radius, as Bryce says. Calculation 1.But that force exceeds the friction between the tire and the ground, then the tire will slip. Calculation 2, with a bit of trig.The opposing force is the gravitational force on the 100 kg load plus the frictional force. The gravitational force is easy, more trig.But what is the frictional force on the 100 kg object? If the object is a block of ice sliding on an slope of ice, then it is low. If the object is a square block of rough granite sliding on a slope of rough granite, then it is high.In other words, you haven’t told us enough to answer the question.
When a tree falls, does the top of the tree exceed the speed of something else that falls from that same height?
Short answer? It depends.Long answer?If I neglect air resistance, I can approximate the scenario to a beam (length 2r) with the weight force acting at the centre of gravity (r metres from the base). Therefore, since the base of the tree acts as a pivot, the top of the tree will accelerate faster than the centre of gravity.Let's assume the torque at any instant is constant along the tree.For a deflection from the vertical, θ, the weight vector perpendicular to the tree is:[math]F=mg\sin{\theta}[/math]Therefore, the torque at the centre of gravity is:[math]T = Fr = mgr \sin{\theta}[/math]We also know that the torque is equivalent to:[math]T = I \alpha = mr^2\alpha[/math]We've assumed here that the tree can be approximated to a point mass at its centre of gravity. This allows us to substitute the mr^2 term for I.Solving this gives:[math]mgr \sin{\theta} = mr^2\alpha[/math][math]g\sin{\theta} = r\alpha[/math][math]\alpha = \frac{g\sin{\theta}}{r}[/math]Angular acceleration, denoted alpha, is constant along the tree. Since the linear acceleration, a, is equal to the product of the angular acceleration and the distance from the pivot point. The acceleration of the top of the tree is therefore:[math]L=2r[/math][math]a = 2r\alpha = 2g\sin{\theta}[/math]The acceleration in the vertical direction is found using trigonometry:[math]y = a\sin{\theta}=2g\sin^2{\theta}[/math]In other words. When:θ < 45°, y < g;θ = 45°, y = g;θ > 45°, y > g.Solving this non-linear differential equation isn't straight forward but if you do it right, you would find that the time taken for the tree to impact with the ground is dependant on the tree's initial angle. Therefore, the starting angle determines if the tree will hit the ground before the object in linear freefall.If that angle is at or greater than 45°, you can be sure that the tree top will definitely hit the ground first.
How do you decide the pivot point when calculating torque?
204 m is LONGER THAN THE LOG, so you should immediately discard that answer. Be careful about blindly applying formulas; make sure the resulting answer makes sense inside the physical system.Here’s how you can envision the problem:The downward force of a 75 kg person on earth will be 75 kg * 9.8 [math]m/s^2 [/math]= 735 NThe log itself has a weight of 420 kg * 9.8 [math]m/s^2[/math] = 4116 NPresume the load is evenly held at both ends, 2058 N each (4116/ 2)The near end of the crossing has no listed limit to the force it can bear. However, the limit force at the other edge is 2650 N, meaning it can stand an extra 592 N (2650 minus 2058) only before collapsing.If Professor Jones stands on the weakened end, it will immediately collapse under his exerted force of 735 N.So, can he get close enough to the end to jump?If he stands in the middle of the log (18m), he will exert half his weight (735 /2 = 367.5 N) at each end of the log. So in the middle of the log, all is cool. So, what percentage of 735 N will equal the critical crossover point of 592 N?735 / x = 592x = 735/592 = 1.24,1/1.24 = 0.805So when he gets 80.5% of the way across the log, it will collapse. 80.5% of 36 m is about 29 m.In conclusion, if he can jump 7 m, he should cross the log.Note that the all time long jump record is 8.95 m. A 7 m jump is possible for a well trained athlete, but ill advised for an aging archaeologist on an unstable surface, barring the fortuitous appearance of a hanging jungle vine.
Physics torque question, Help please.?
There is a vertical upward component of torque supplied by the cable, that has a constant maximum. What you're trying to find out is the point at which the vertical downward total of the torque exceeds this maximum, breaking the cable. Start with force: the cable is at 37 degrees and has a breaking strain of 8000N. The vertical component of that will be given by 8000 sin 37 = 4815 N. Convert this into a torque with T = F*r = 4814*5 = 24,073 nm That's the total 'upward' torque that can be provided by the cable, just before it breaks. The 'downward' torque due to the man will just be given by T = F*r, but the force is his mass times g, so it's T =m*g*r = 95*9.8*r = 931r nm (we are trying to solve for this particular value of 'r' - it's *not* the 5m - maybe make it r(man) or something to remind you. The 'downward' torque due to the mass of the beam itself will be its weight force, mg times its radius... but since it's a solid beam the radius is neither the 5m where the cable is attached or its 8 m total length. It's been a while since I did these - can you treat it as though the radius is the centre of mass, 4m? Or do you need to do an integral along the whole length? Give me a moment and I'll do it on paper... Right, both ways yield the same answer, so we *can* just say that the 'downward' torque due to the beam is T =m*g*r (where r = 4 m) = 23,520 nm. Now, the cable can support 24,073 m, and the beam requires 23,520 of that to support it, so subtract the latter: 24,073-23,520 = 553 nm This needs to be equal to the 'downward' torque due to the man to give the point at which the cable breaks: 553 = 931r Rearranging, r = 0.59 m That is, if the man walks further out the beam than 59 cm, 0.59 m, the cable will break