What is the maximum value for w1? Under this maximum weight, what is FBy, the vertical component of the support's reaction force at point B?
As shown, a uniform beam of length d = 1.50m and mass 15.5kg is attached to a wall with a pin at point B. (Figure 3) A cable attached at point A supports the beam. The beam supports a distributed weight w2 = 395N/m . If the support cable can sustain a maximum tension of 1250N , what is the maximum value for w1? Under this maximum weight, what is FBy, the vertical component of the support's reaction force at point B?
A uniform horizontal bar of length L = 5 m and weight 207 N is pinned to a vertical wall and supported?
a) Resolve the torques about point A. We have the vertical sin(θ) component of the tension rotating counter-clockwise, the weight of the bar rotating clockwise, and the weight of the box rotating clockwise. Recall that gravity acts at the center of mass, which is located at L / 2. Thus, balancing these gives: T (tension) = T (bar) + T (box) L T sin(θ) = (L / 2) Wb + x Wm x = [L T sin(θ) - (L / 2) Wb] / [Wm] Where Wb is the weight of the bar, Wm is the weight of the mass, and T is the maximum value of the tension possible. Subbing in the numbers: x_{max} = [(5 m)(503 N)(sin(15⁰)) - (1 / 2)(5 m)(207 N)] / [361 N] x_{max} ≈ 0.369 m b) Balance the forces in the horizontal direction: F (horizontal bar) = F (horizontal tension) F (horizontal bar) = T cos(θ) F (horizontal bar) = (503 N) cos(15⁰) F (horizontal bar) ≈ 486 N c) Balance the forces in the vertical direction: F (vertical bar) + F (vertical tension) + F (bar weight) + F (mass weight) = 0 F (vertical bar) + T sin(θ) - Wb - Wm = 0 F(vertical bar) + (503 N) sin(15⁰) - 207 N - 361 N = 0 F (vertical bar) ≈ 438 N Done!
A particle has an initial velocity of 3i + 4j and an acceleration of 0.4i + 0.3j. What is the speed after 10s?
Initial Velocity is the velocity at which motion begins. It is velocity at time interval t = 0. It is commonly denoted by u.Initial Velocity formula is used to find the initial velocity of the body if some of the quantities (like acceleration and time interval) are known. Initial velocity is expressed in meter per second (m/s). The initial velocity is given byu = v - atWhere, v = Final Velocity, t = time taken, and a = accelerationu = 3 i + 4 ja = 0.4 i + 0.3 jt = 10 sUsing initial velocity formula3 i + 4 j = v - (0.4 i + 0.3 j) 10v = 7 ( i + j)Velocity is the vector quantity which is defined by both magnitude and direction but speed is scalar quantity which is defined by only magnitude.The magnitude of v = ([math]7^{2} + 7^{2})^{1/2} = 7 * (2)^{1/2}[/math]So the value of speed is [math]7 * (2)^{1/2}[/math]
Classical Physics: A bomb is dropped from a plane flying horizontal with speed of 720km/hr, at an altitude of 980m. After what time the bomb will hit the ground?
Using the equation y=(v*t) + (1/2)a(t)^2 with 980m for y, 0 for v as the initial velocity was entirely horizontal, and 9.8m/s^2 for a as it is very close to the acceleration near the surface of Earth. We come to 980 meters = 4.9t^2 meters per second^2 which when further simplified yields 200 s^2 = t^2. T then must equal 10*2^(-2) seconds or about 14.14 seconds.Using this time we can also calculate how far horizontally the bomb traveled by taking 720 kilometers per hour, multiplying it by a 1000 to convert it to meters per hour and dividing it by 3600 to convert it to meters per second. This yields the value of 200 m/s which when multiplied with the time we found earlier results in a horizontal distance of around 2826.427 meters (if one multiplies by the exact value of time found earlier). So in summary the bomb fell 980m in around 14.14 seconds and traveled horizontally at a velocity of 200 m/s to a total horizontal distance of 2828.427m.Edit: This assumes no air resistance which is not entirely realistic but in the case of a bomb that would likely amount to at most a 2 second drop increase and maybe a reduction in horizontal distance of at the very greatest around 2000m without taking wind speeds into account. Wind speeds might reduce the distance by a maximum value of an additional 200 to 300m. Keep in mind this is a fairly critical evaluation of the factors at play and is a bit of an overestimate of what the actual values might be.Hope this was helpful! Edit 2: By the way, if you are dropping a fission or fission/part fusion weapon you would want to be much higher than that and also turning away your plane (if you are using a plane) so as to escape before it detonates, besides the fact that most fission bombs are designed to be slowly dropped by a parachute and detonated in air to allow for the greatest destructive potential and the safe escape of the bomber and crew.
A block of mass 3kg is placed on a rough surface. The coefficient of static friction between two surfaces is 0.2 then what is the minimum horizontal force required to move the block?
As stated, there is not sufficient information to answer.What is the acceleration due to gravity whereas the experiment is performed?If you can assume 1 Earth standard gravity, the the block is pressed down onto the surface by a force of 9.8 * 3 N. The coefficient of friction presents a retarding force of 0.2 * 29.4, or 5.88N
How to explain the slope of value k by hookes law?
Find the equilibrium position for a given mass on the spring. pull the mass down a measured distance x. Release the mass and measure the period T of the up and down motion. Repeat two other strecthes of the spring and two other masses. Make a plot of m vs. T^2 on a graph does the slopeof the graph equal k? I did all the graphs and stuff but I dont know how to explain the last part "does the slopeof the graph equal k?" I have already found k.