A wheelbarrow full of soil is being pulled to the right. Below the wheelbarrow is a free body diagram with 4 force vectors. The first vector is pointing downward, labeled F Subscript g Baseline = negative 800 N. The second vector is pointing left, labeled F Subscript f Baseline = negative 400 N. The third vector is pointing upward, labeled F Subscript N Baseline = 800 N. The fourth vector is pointing right labeled F Subscript t Baseline = 250 N. The up and down vectors are the same length. The vector pointing right is shorter than the vector pointing left.
Based on the free-body diagram, the net force acting on this wheelbarrow is
N.

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jossephpineda124575 jossephpineda124575 · Answer 1 · 2019-09-27T22:15:34+02:00

Answer:

800n 400n 250n 800n i think is 300n

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sp10925 sp10925 · Answer 2 · 2022-07-14T08:30:08+02:00

The net force acting on this wheelbarrow is 650 N.

When two or more forces are acting on the system of objects, then the to attain equilibrium, net force must be zero.

Given is a wheelbarrow full of soil is being pulled to the right. The wheelbarrow is a free body diagram with 4 force vectors.

For net force in X direction, ∑Fx = 0

So, ∑Fx = Ft -Ff =0

∑Fx = 250 -(-400) = 650 N

For net force in Y direction, ∑F = 0

The vectors are of same length and magnitude, so they will cancel each other.

So, the net force in Y direction is zero.

Net force will be F= 650 N.

Thus, the net force acting on this wheelbarrow is 650 N.

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