The diagram below shows a block of mass on a frictionless horizontal surface, as

Question

The diagram below shows a block of mass on a frictionless horizontal surface, as seen from above. Three forces of magnitudes ,F1=4N angle 25,F2=6N 180,F3=8N 325 are applied to the block, initially at rest on the surface, at angles given. In this problem, you will determine the resultant (total) force vector from the combination of the three individual force vectors. All angles should be measured counterclockwise from the positive x axis (i.e., all angles are positive). Calculate magnitude of resultant force acting on block , Fr =F1 + F2 + F3 What angle does Fr make with positive x axis

Explanation / Answer

Hi, F1=4N angle 25,F2=6N 180,F3=8N 325 are applied to the block The horizontal component of resultant force acting on the block = Fx =                                     = F1x + F2x + F3x                                     = (4*Cos(25)) + (6*Cos(180)) + (8*Cos(325))                                     = 4.1785N Similarly the vertical component of resultant force acting on the block = Fy                                     = F1y + F2y + F3y                                     = (4*Sin(25)) + (6*Sin(180)) + (8*Sin(325))                                     = -2.8981N Hence the magnitude of resultant force = F = sqrt(Fx^2 + Fy^2)                                              = 5.0852N and in the direction making an anlge ? = arctan(Fy/Fx) = -34.74 deg with +ve x-axis
                                                   = 325.26 deg with +ve x-axis counter-clockwise.
We know F = M * a. Hence acceleration of the block = a = F / M                                                                               = (5.0852N / M) m/s^2 substitute the value of M if know to get the final value. The direction of acceleration will be same as the direction of force i.e., 325.26 deg with +ve x-axis counter-clockwise.
The distance the mass moves in 5sec = s = ut + (1/2)at^2                      = 0 + (0.5 * (5.0852 / M) * 5^2) = (63.565/M) m. The magnitude of its velocity at this point = v = u + at = 0 + (5.0852 / M) * 5                                                                      = (25.426/M) m/s         The velocity will also be along the same direction as the acceleration is acting and hence in the same direction as the force is acting. Hope this helps you. The velocity will also be along the same direction as the acceleration is acting and hence in the same direction as the force is acting. Hope this helps you.

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