Question

icw 17a a 250 lb block is subjexted to a horizontal force p. the coefficient of friction between the block and surface is 0.2 determine the force P required to start moving the block up the incline

Answer 1

424.773N

Step-by-step explanation:

Taking the sum of force along the incline. The forces acting along the incline is the moving force P and the frictional force Ff

P = mgsintheta

Ff = nR = nmgcostheta

n is the coefficient of friction

R is the normal reaction

m is the mass of the block

g is the acceleration due to gravity

`\sum Fx = 0\\P-Ff = 0\\mgsin\theta - \mu mgcos\theta = 0\\mgsin\theta = \mu mgcos\theta\\P = \mu mgcos\theta\\P = (0.2)250(9.81)cos30^0\\P = 490.5cos30^0\\P = 490.5(0.8660)\\P = 424.773N\\`

Hence the force P required to start moving the block up the incline is 424.773N