Recent content by bobmarly12345

Homework Statement i had a solid rectangular beam of max stress equalling 1.65x10^6N/m² a new beam, a hollow rectangular beam is made to reduce weight, it has the same maximum bending stress 1.65x10^6N/m² but this time has a horizontal dimension of 0.2m and a constant wall thickness of 0.02...

brilliant thanks for your help! :)

yeah just divide by 9.81 i believe ?

oh yeah i see now why its only the vertical component, so just 510xsin25=215.5 that's all?

oh yes i forgot about the effect of gravity on the 52Kg block. ok i understand to a point about the mass of block k , so would the mass of block k be HN x (cos25+sin25) ? then divide by gravitational constant to get the Kg value ?

well the question asks me to work out the mass of K

Homework Statement An assembly shown in the picture, GH , HK, HNL are three cables. the mass of box L is 52Kg, a) determine the mass of box K in order to make angle theta as 25 degrees Homework Equations FX=0 FY=0 The Attempt at a Solution right i started with Fhn x...

Homework Statement can anyone show me what the free body diagram of this would look like? Thankyou

Homework Statement The instantaneous configuration of a slider crank mechanism has a crank GH 10cm long, the connecting rod HP is 50cm. The crank makes an angle of 60 degree with the inner dead centre position and is rotating at 110 rev/min. Determine the velocity of the piston P and the...

Homework Statement a golf ball is struck at an angle of 30 degree with respect to the horizontal, at a velocity of 11m/sec. the golf course is perfectly level and the golf ball stops instantaneously when it hits the ground. a) determine the horizontal distance traveled by the golf ball b)...

Homework Statement An assembly shown in the picture, GH , HK, HNL are three cables. the mass of box L is 52Kg, a) determine the mass of box K in order to make angle theta as 25 degrees Homework Equations Fx=0 Fy=0 The Attempt at a Solution Fhn x cos(25) = Fhg x cos(0) fhn x...

oh of course yes! thank you for your help :)

so from that then would it follow that (I1.u1) +(I2.u2) = (I1+I2).V2,,,,,, V2=(I1.u1) +(I2.u2)/(I1+I2). therefore, I1=40X0.111m²=0.49284 & 10x0.1m²=0.1 then v2=9.98rev/s ? Would you agree with that or am i going in the wrong direction :)

yeah I've got that part but how does the moment of inertia from that equation help me with finding the impact of the engine and drum? do i replace the mass in the impact before and after with I=M.R² so it would be (I1.u1) +(I2.u2) = (I1.v1)+(I2.v2) or do i use a different formula all together? I...

would this formula work, [ -rw(cos(60)+((sin2(60)/(2xsqrt(n²-sin²(60)))) ] where n = L/R i get a negative value when i use it, the piston is moving in the negative direction.